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:

Two ways.. Open the MOD.. |2x+5|=|3x-2|.... Square both sides... \((2x+5)^2=(3x-2)^2.......4x^2+20x+25=9x^2-12x+4....5x^2-32x-21=0.........(5x+3)(x-7)=0\).. So x can be -3/5 or 7...

Second is substitution.. Start with the central value.. x=-3/5.. |2*-3/5+5|=|3*-3/5-2|........|5-6/5|=|-9/5-2|=19/5=19/5... Correct

When you open the mod take care to solve for both positive and negative situations i.e \(±\)

Here \(|2x+5|=|3x-2|\). Now lets decide to open the LHS mod

so we will have \(2x+5=±|3x-2| => 2x+5=|3x-2|\) and \(2x+5=-|3x-2|\)

Case 1: \(|3x-2|=2x+5\), again we have a mod so we will have further two cases \(3x-2 =±(2x+5)\)

so \(3x-2=2x+5 => x=7\)

and \(3x-2=-(2x+5) => x= -\frac{3}{5}\)

Case 2: \(-|3x-2|=2x+5\) or \(|3x-2|=-2x-5\), again we have a mod so we will have further two cases \(3x-2=±(-2x-5)\)

so \(3x-2=-2x-5 => x=-\frac{3}{5}\)

and \(3x-2 = 2x+5 => x=7\)

So finally we have \(x=7\) or \(\frac{-3}{5}\)

Hence Option C

Alternatively as Chetan had explained, you can square both sides to remove the mod. the only challenge in this method can be, if the equation is complex then it will lead to further complexity.

He simply opened LHS modulus to take two cases of positive and negative terms of RHS. But interestingly he probably assumed that LHS will be always be positive ie no scope for -2x-5

He simply opened LHS modulus to take two cases of positive and negative terms of RHS. But interestingly he probably assumed that LHS will be always be positive ie no scope for -2x-5

To be fail-safe, in my opinion, when we remove mod we need to consider both positive and negative scenarios for each expression

For this question squaring works best because the equation is linear and has only 1 variable x. Squaring should result in a simple quadratic equation. If the equation had been any other polynomial or had contained more than one variable, then squaring method would have been cumbersome.

Finally, you can always assume two scenarios for variables -

Case 1: if 2x+5>0 or x>-5/2, then |3x-2|=2x+5, solve this to get two values of x that satisfy x>-5/2

Case 2: if 2x+5<0 or x<-5/2, then |3x-2|=-(2x+5).....{because if x<0, then |x|=-x}, solve this to get two values of x that satisfy x<-5/2

then combine the values of x derived from the above two cases.

Would request other experts to chip-in with their thoughts

He simply opened LHS modulus to take two cases of positive and negative terms of RHS. But interestingly he probably assumed that LHS will be always be positive ie no scope for -2x-5

To be fail-safe, in my opinion, when we remove mod we need to consider both positive and negative scenarios for each expression

For this question squaring works best because the equation is linear and has only 1 variable x. Squaring should result in a simple quadratic equation. If the equation had been any other polynomial or had contained more than one variable, then squaring method would have been cumbersome.

Finally, you can always assume two scenarios for variables -

Case 1: if 2x+5>0 or x>-5/2, then |3x-2|=2x+5, solve this to get two values of x that satisfy x>-5/2

Case 2: if 2x+5<0 or x<-5/2, then |3x-2|=-(2x+5).....{because if x<0, then |x|=-x}, solve this to get two values of x that satisfy x<-5/2

then combine the values of x derived from the above two cases.

Would request other experts to chip-in with their thoughts

I am not an expert but would like to share my thoughts here.

When we have modulus on both side of the equality (=) sign and when we are trying the cases approach, we only need to consider two cases. First, both sides have same sign. Second, different sign on both sides.

One may try the other two cases as well, but the results will be the same as above.

Hope that helps. Let me know if you want more details.