Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

solving these equations you will have x=2 or x = 2 then (1) is sufficient. X = 2

(2) 5x+3 = 2x+9 or 5x+3 = -(2x+9)

3x = 6 or 7x = -12 we will have two values for x. (2) is not sufficient

Would pick (A)

I agree A is right. But do not neglect 2nd statement just because it gave 2 answers.

5x+3 = -(2x+ 9) only when 5x+3 <0 so the answer obtained 7x=-12 must satisfy this domain. In this case it is satisfying thus making it insufficient.

Imagine a case when the 2nd solution does not satisfy the domain, in that case it will be sufficient. So do not neglect the statement just because it gave two answers.
_________________

As a module is always positive, hence we can say that x-2>0 If x-2>0 then it implies x>2. But our answer is showing that x=2

What am I missing here??

Red part is not correct: absolute value is never negative (absolute value \(\geq{0}\)).

Step-by-step solution:

(1) \(|6 - 3x| = x - 2\) --> as LHS (left hand side) is absolute value, which is never negative (absolute value \(\geq{0}\)), RHS also must be \(\geq{0}\) --> \(x-2\geq{0}\) --> \(x\geq{2}\). For \(x\geq{2}\), \(|6 - 3x|=-6+3x\), so \(|6 - 3x| = x - 2\) becomes \(-6+3x=x-2\) --> \(x=2\). Sufficient.

(2) \(|5x + 3| = 2x + 9\). (note that by the same logic as above \(x\geq-\frac{9}{2}=-4.5\)) One check point \(x=-\frac{3}{5}\) (check point(s) the value of x when absolute value equals to zero).

\(x\leq{-\frac{3}{5}}\) --> \(-5x-3=2x+9\) --> \(x=-\frac{12}{7}\) - valid solution as \(-\frac{9}{2}<-\frac{12}{7}<-\frac{3}{5}\);

\(x>{-\frac{3}{5}}\) --> \(5x+3=2x+9\) --> \(x=2\) - valid solution as \(-\frac{9}{2}<-\frac{3}{5}<2\);

As a module is always positive, hence we can say that x-2>0 If x-2>0 then it implies x>2. But our answer is showing that x=2

What am I missing here??

Red part is not correct: absolute value is never negative (absolute value \(\geq{0}\)).

Step-by-step solution:

(1) \(|6 - 3x| = x - 2\) --> as LHS (left hand side) is absolute value, which is never negative (absolute value \(\geq{0}\)), RHS also must be \(\geq{0}\) --> \(x-2\geq{0}\) --> \(x\geq{2}\). For \(x\geq{2}\), \(|6 - 3x|=-6+3x\), so \(|6 - 3x| = x - 2\) becomes \(-6+3x=x-2\) --> \(x=2\). Sufficient.

(2) \(|5x + 3| = 2x + 9\). (note that by the same logic as above \(x\geq-\frac{9}{2}=-4.5\)) One check point \(x=-\frac{3}{5}\) (check point(s) the value of x when absolute value equals to zero).

\(x\leq{-\frac{3}{5}}\) --> \(-5x-3=2x+9\) --> \(x=-\frac{12}{7}\) - valid solution as \(-\frac{9}{2}<-\frac{12}{7}<-\frac{3}{5}\);

\(x>{-\frac{3}{5}}\) --> \(5x+3=2x+9\) --> \(x=2\) - valid solution as \(-\frac{9}{2}<-\frac{3}{5}<2\);

Two valid solutions. Not sufficient.

Answer: A.

Hope it helps.

I am not satisfied with the solution.

How do we solve modulus, we take two cases, 1)if (6-3x)>0 then given equation comes out to be 6-3x=x-2 which gives x=2 ,but we are considering the case when 6-3x>0 or x<2, hence this soln x=2 is not possible.

2) When we consider (6-3x)<0 then equation comes out to be -6+3x= x-2 which again gives x=2 , but this time we are considering 6-3x<0 or x>2, hence this soln x=2 is not feasible in this case also.

As a module is always positive, hence we can say that x-2>0 If x-2>0 then it implies x>2. But our answer is showing that x=2

What am I missing here??

Red part is not correct: absolute value is never negative (absolute value \(\geq{0}\)).

Step-by-step solution:

(1) \(|6 - 3x| = x - 2\) --> as LHS (left hand side) is absolute value, which is never negative (absolute value \(\geq{0}\)), RHS also must be \(\geq{0}\) --> \(x-2\geq{0}\) --> \(x\geq{2}\). For \(x\geq{2}\), \(|6 - 3x|=-6+3x\), so \(|6 - 3x| = x - 2\) becomes \(-6+3x=x-2\) --> \(x=2\). Sufficient.

(2) \(|5x + 3| = 2x + 9\). (note that by the same logic as above \(x\geq-\frac{9}{2}=-4.5\)) One check point \(x=-\frac{3}{5}\) (check point(s) the value of x when absolute value equals to zero).

\(x\leq{-\frac{3}{5}}\) --> \(-5x-3=2x+9\) --> \(x=-\frac{12}{7}\) - valid solution as \(-\frac{9}{2}<-\frac{12}{7}<-\frac{3}{5}\);

\(x>{-\frac{3}{5}}\) --> \(5x+3=2x+9\) --> \(x=2\) - valid solution as \(-\frac{9}{2}<-\frac{3}{5}<2\);

Two valid solutions. Not sufficient.

Answer: A.

Hope it helps.

I am not satisfied with the solution.

How do we solve modulus, we take two cases, 1)if (6-3x)>0 then given equation comes out to be 6-3x=x-2 which gives x=2 ,but we are considering the case when 6-3x>0 or x<2, hence this soln x=2 is not possible.

2) When we consider (6-3x)<0 then equation comes out to be -6+3x= x-2 which again gives x=2 , but this time we are considering 6-3x<0 or x>2, hence this soln x=2 is not feasible in this case also.

So according to me answer is E.

Note that OA for this questio is A, not E.

Next, to check whether x=2 is a root of |6 - 3x| = x - 2 just substitute x=2 into it: |6 - 3*2| = 0 = 2 - 2.

It seems that I have solved some abs. value problems by finding what makes the equation in the abs. value sign = to zero, so for #2, why do we find what makes the equation outside the abs. value sign = to zero? In other words...

It seems that I have solved some abs. value problems by finding what makes the equation in the abs. value sign = to zero, so for #2, why do we find what makes the equation outside the abs. value sign = to zero? In other words...

We have \(|6 - 3x| = x - 2\). The left hand side (|6 - 3x|) is an absolute value. The absolute value is always more than or equal to 0, thus the right hand side (x - 2) must also be \(\geq{0}\) --> \(x-2\geq{0}\) --> \(x\geq{2}\). When \(x\geq{2}\), \(6 - 3x\leq{0}\), thus \(|6 - 3x|=-(6-3x)=-6+3x\), so \(|6 - 3x| = x - 2\) becomes \(-6+3x=x-2\) --> \(x=2\). Sufficient.

Similar solution from in both scenario. Hence sufficient.

Statement 2:

Similar process and we will get x= 12/7 and 2. Two different solutions. hence Insufficient.

Ans: A

Atal Pandit

Yes, we can do this way too.

Typo: |5x + 3| = 2x + 9 --> x=-12/7 or x=2.

Hello Bunuel,

I followed the same approach but looking at solution given by you, It is actually not required . I have seen some question types where in we just check for one condition (X>/0 or X<0) and we can conclude the answer (Like you just did in this one)

Are there particular question type where in your approach to this question will work. For ex | Some expression in x| = x- some no.

Again since LHS >/0, we can say x-some no >/ -----> x>/ some no.

This can be helpful in saving time by avoiding checking expression for 2 possible case of x (<0,>0)

Thanks Mridul
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

I followed the same approach but looking at solution given by you, It is actually not required . I have seen some question types where in we just check for one condition (X>/0 or X<0) and we can conclude the answer (Like you just did in this one)

Are there particular question type where in your approach to this question will work. For ex | Some expression in x| = x- some no.

Again since LHS >/0, we can say x-some no >/ -----> x>/ some no.

This can be helpful in saving time by avoiding checking expression for 2 possible case of x (<0,>0)

Thanks Mridul

Yes, in such kind of questions this approach might be helpful.
_________________

I just wanted to make sure the way I solved this problem is correct:

What is the value of x?

1. |6-3x|=x-2 2. |5x+3|=2x+9

1. 6-3x is positive when x<2. For example, if x=1 then 6-3x = 6-3(1) = 3. 6-3x is negative when x>2. For example, if x=3 then 6-3(x) = 6-3(3) = -3.

For positive cases: 6-3x=x-2 ==> x=2 For negative cases: -6+3x=x-2 ==> x=2

My issue here is that for the positive case x must be less than 2 and for the negative case x must be greater than 2 but for both cases here, I got x =2. The answer is sufficient, but I thought that for the cases to be sufficient the results had to be in line with what x= (for example, for the positive case here, x=2 but before I found that X must be LESS than 2 for 6-3x to be positive)

Could someone explain? I see why #2 fails (because for the positive case, 2 is greater than -3/5 but for the negative case, -7/12 is NOT greater than -3/5 but I am a bit lost with #1)

I just wanted to make sure the way I solved this problem is correct:

What is the value of x?

1. |6-3x|=x-2 2. |5x+3|=2x+9

1. 6-3x is positive when x<2. For example, if x=1 then 6-3x = 6-3(1) = 3. 6-3x is negative when x>2. For example, if x=3 then 6-3(x) = 6-3(3) = -3.

For positive cases: 6-3x=x-2 ==> x=2 For negative cases: -6+3x=x-2 ==> x=2

My issue here is that for the positive case x must be less than 2 and for the negative case x must be greater than 2 but for both cases here, I got x =2. The answer is sufficient, but I thought that for the cases to be sufficient the results had to be in line with what x= (for example, for the positive case here, x=2 but before I found that X must be LESS than 2 for 6-3x to be positive)

Could someone explain? I see why #2 fails (because for the positive case, 2 is greater than -3/5 but for the negative case, -7/12 is NOT greater than -3/5 but I am a bit lost with #1)

Thanks!

You have a point here.

When you study a function, an abs value, you correctly define where is positive or negative \(x>a\) positve for example \(x<a\) negative for example As you see the value \(a\) is not contained in neither interval... How do we solve this? Just put an \(=\) sign under one of the above inequalities, so \(x\geq{}a\) and \(x<a\) - or \(x>a\) and \(x\leq{}a\). Where you put the sign, it doesn not matter, the result will not change.

So just rewrite your equation into

For positive cases (\(x\leq{2}\)): 6-3x=x-2 ==> x=2 VALID For negative cases (\(x>2\)): -6+3x=x-2 ==> x=2

and you have the result! Nothing changes, everything you did is absolutely correct, just put an \(=\) sign in one of the intervals Hope it's clear
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Ok...I see that. But isn't the whole point to test cases where 6-3x is positive and negative? if X=2 then 6-3x is neither positive or negative, it's just zero.

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...