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 350,000 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:

Statement 1: x = 2 For (x+3) >= 0 x >= -3 [Not a solution as it doesn't satisfies the condition] x+3=4x-3 x = 2

For (x+3) < 0 x < -3 [Not a solution as it doesn't satisfies the condition] -(x+3) = 4x-3 x = 0 [Not a solution as it doesn't satisfies the condition]

Statement 2: x = 2 For (x+1) >= 0 x >= -1 [Not a solution as it doesn't satisfies the condition] x+1 = 2x-1 x = 2

For (x+1) < 0 x < -1 [Not a solution as it doesn't satisfies the condition] -(x+1) = 2x - 1 x = 0 [Not a solution as it doesn't satisfies the condition]

To save time, we should ignore all -ve cases of x as LHS is +ve and equal to RHS. So, for RHS to be positive x must be +ve. _________________

The three most significant times in your life are: 1. When you fall in love 2. The birth of your first child 3. When you prepare for your GMAT

stmt1 ... when |x+3|<0 -(x+3) = 4x-3 5x=0,x=0 ....

Here is a typical mistake with moduli. You forgot to check your answer. It is possible to use -(x+3) only if |x+3|<0 ! But at x=0, |0+3| > 0. _________________

Stmt 1:|x + 3| = 4x – 3 Two cases: a. x+3=4x-3 ,x=0 b. -(x+3)=4x-3 ,x=0.

Both cases gives us x=0 hence SUFF.

Stmt 2:|x + 1| = 2x – 1 Again two cases: a.x+1=2x-1 x-2x=-1-1 x=2

b. -(x+1)=2x-1 x=0 Both case gives us diff. values hence insuff. IMO: A

Please point out my flaw.

First you made a mistake in calculation: see the red part, solution there is x=2. Second, which is more important the way of solving, from my point of view, is incorrect.

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.

(2) \(|x+1|=2x-1\) --> the same here --> \(2x-1\geq{0}\) --> \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient.

So you see that you don't even need to find exact value(s) of \(x\) to answer the question.

Stmt (1) First we squire both sides of the equality: |x + 3| = 4x – 3 x^2+6x+9=16x^2-24x+9 15x^2-30x=0 divide both sides to 15 and we have: x^2-2x=0 x(x-2)=0 x=0 but x=0 is not a root of the equation since |0 + 3| is not equal to 4*0 – 3= -3 because absolute value cannot equal to digit below zero x-2=0 x=2 is a root of the equality, since |2 + 3| = 4*2 – 3=> 5 both sides hence (1) is sufficient

stmt (2) |x + 1| = 2x – 1 squire both sides again and we have: x^2+2x+1=4x^2-4x+1 3x^2-6x=0, divide both sides to 3 and have: X^2-2x=0 x(x-2)=0 x=0 but x=0 is not a root of the equation since |0 + 1| is not equal to 2*0 – 1= -1 because absolute value cannot equal to digit below zero x-2=0 x=2 is a root of the equality, since |2 + 1| = 2*2 – 1=> 3 both sides

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. _________________

(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.

(2) \(|x+1|=2x-1\) --> the same here --> \(2x-1\geq{0}\) --> \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient.

So you see that you don't even need to find exact value(s) of \(x\) to answer the question.