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:

Re: [tricky one] Is x > 0? [#permalink]
04 Jun 2013, 07:42

Expert's post

1

This post was BOOKMARKED

\(|x+3|=4x-3\). Check point is at \(x=-3\) (check point, is the value of x for which the value of an expression in modulus equals to zero).

When \(x\leq{-3}\), then \(x+3<0\), thus \(|x+3|=-(x+3)\). So, in this case we have \(-(x+3)=4x-3\) --> \(x=0\) --> discard this solution since we consider the range when \(x\leq{-3}\).

When \(x>{-3}\), then \(x+3>0\), thus \(|x+3|=x+3\). So, in this case we have \(x+3=4x-3\) --> \(x=2\) --> since \(x=2>-3\), then this solution is valid.

So, we have that \(|x+3|=4x-3\) has only one root: \(x=2\).

Re: [tricky one] Is x > 0? [#permalink]
04 Jun 2013, 08:10

Bunuel wrote:

\(|x+3|=4x-3\). Check point is at \(x=-3\) (check point, is the value of x for which the value of an expression in modulus equals to zero).

When \(x\leq{-3}\), then \(x+3<0\), thus \(|x+3|=-(x+3)\). So, in this case we have \(-(x+3)=4x-3\) --> \(x=0\) --> discard this solution since we consider the range when \(x\leq{-3}\).

When \(x>{-3}\), then \(x+3>0\), thus \(|x+3|=x+3\). So, in this case we have \(x+3=4x-3\) --> \(x=2\) --> since \(x=2>-3\), then this solution is valid.

So, we have that \(|x+3|=4x-3\) has only one root: \(x=2\).

Re: [tricky one] Is x > 0? [#permalink]
04 Jun 2013, 08:13

1

This post received KUDOS

Expert's post

WholeLottaLove wrote:

Bunuel wrote:

\(|x+3|=4x-3\). Check point is at \(x=-3\) (check point, is the value of x for which the value of an expression in modulus equals to zero).

When \(x\leq{-3}\), then \(x+3<0\), thus \(|x+3|=-(x+3)\). So, in this case we have \(-(x+3)=4x-3\) --> \(x=0\) --> discard this solution since we consider the range when \(x\leq{-3}\).

When \(x>{-3}\), then \(x+3>0\), thus \(|x+3|=x+3\). So, in this case we have \(x+3=4x-3\) --> \(x=2\) --> since \(x=2>-3\), then this solution is valid.

So, we have that \(|x+3|=4x-3\) has only one root: \(x=2\).

Hope it's clear.

Why do we discard that solution?

We consider the range \(x\leq{-3}\). x=0 is out of this range.

Re: [tricky one] Is x > 0? [#permalink]
04 Jun 2013, 10:25

I think I got it. Thanks!

Bunuel wrote:

WholeLottaLove wrote:

Bunuel wrote:

\(|x+3|=4x-3\). Check point is at \(x=-3\) (check point, is the value of x for which the value of an expression in modulus equals to zero).

When \(x\leq{-3}\), then \(x+3<0\), thus \(|x+3|=-(x+3)\). So, in this case we have \(-(x+3)=4x-3\) --> \(x=0\) --> discard this solution since we consider the range when \(x\leq{-3}\).

When \(x>{-3}\), then \(x+3>0\), thus \(|x+3|=x+3\). So, in this case we have \(x+3=4x-3\) --> \(x=2\) --> since \(x=2>-3\), then this solution is valid.

So, we have that \(|x+3|=4x-3\) has only one root: \(x=2\).

Hope it's clear.

Why do we discard that solution?

We consider the range \(x\leq{-3}\). x=0 is out of this range.

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
17 Jun 2013, 22:45

Hi Bunnel, I've been struggling a lot in inequalities.. I've a question pertaining to solution provided by saxenaashi ...

1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard[Understood this part]2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution )[Did not understand, Why -ve sign is here as the x range is positive]3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)[Understood] Another thing is 3/2(inclusive) is not at all considered in the range in any of the three checkpoints.

What am i missing here and how to solve these questions effectively

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
17 Jun 2013, 23:50

Expert's post

kawan84 wrote:

Hi Bunnel, I've been struggling a lot in inequalities.. I've a question pertaining to solution provided by saxenaashi ...

1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard[Understood this part]2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution )[Did not understand, Why -ve sign is here as the x range is positive]3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)[Understood] Another thing is 3/2(inclusive) is not at all considered in the range in any of the three checkpoints.

What am i missing here and how to solve these questions effectively

Absolute value properties:

When \(x\leq{0}\) then \(|x|=-x\), or more generally when \(some \ expression\leq{0}\) then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(x\geq{0}\) then \(|x|=x\), or more generally when \(some \ expression\geq{0}\) then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

SOLUTION:

We have two transition points for \(|x-3|=|2x-3|\): \(x=\frac{3}{2}\) and \(x=3\). Thus three ranges to check: 1. \(x<\frac{3}{2}\); 2. \(\frac{3}{2}\leq{x}\leq{3}\); 3. \(3<x\)

Note that it does not matter in which range(s) you include the transition points with "=" sign as long you include them.

1. When \(x<\frac{3}{2}\), then \(x-3\) is negative and \(2x-3\) is negative too, thus \(|x-3|=-(x-3)\) and \(|2x-3|=-(2x-3)\).

Therefore for this range \(|x-3|=|2x-3|\) transforms to \(-(x-3)=-(2x-3)\) --> \(x=0\). This solution is OK, since \(x=0\) is in the range we consider (\(x<\frac{3}{2}\)).

2. When \(\frac{3}{2}\leq{x}\leq{3}\), then \(x-3\) is negative and \(2x-3\) is positive, thus \(|x-3|=-(x-3)\) and \(|2x-3|=2x-3\).

Therefore for this range \(|x-3|=|2x-3|\) transforms to \(-(x-3)=2x-3\) --> \(x=2\). This solution is OK, since \(x=2\) is in the range we consider (\(\frac{3}{2}\leq{x}\leq{3}\)).

3. When \(3<x\), then \(x-3\) is positive and \(2x-3\) is positive too, thus \(|x-3|=x-3\) and \(|2x-3|=2x-3\).

Therefore for this range \(|x-3|=|2x-3|\) transforms to \(x-3=2x-3\) --> \(x=0\). This solution is NOT OK, since \(x=0\) is NOT in the range we consider (\(3<x\)).

Thus \(|x-3|=|2x-3|\) has two solutions \(x=0\) and \(x=2\).

Re: [tricky one] Is x > 0? [#permalink]
21 Jun 2013, 23:40

Bunuel wrote:

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

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.

For statement 1, let us square both sides of the equation and remove the modulus sign.

Re: [tricky one] Is x > 0? [#permalink]
22 Jun 2013, 02:43

Expert's post

navigator123 wrote:

Bunuel wrote:

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

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.

For statement 1, let us square both sides of the equation and remove the modulus sign.

EDIT: Apparently you CAN square both sides in #2. I thought you had to know that, for example (x-3) and (2x-3) are positive in order for you to be able to square both sides?

Re: [tricky one] Is x > 0? [#permalink]
27 Jul 2013, 09:54

Bunuel wrote:

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

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.

for (2) why do you start by squaring both sides? What is the rule that you are using to determine that this is the first step?

Re: [tricky one] Is x > 0? [#permalink]
27 Jul 2013, 13:06

You can square both sides of an inequality or equation when there are absolute values involved. Think of it like this:

x^2 = y^2 |x| = |y|

Lets say x^2 = 4. Now, x can be 2 OR -2 because squaring any number positive or negative will result in a number ≥ 0.

kanderoo wrote:

Bunuel wrote:

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

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.

for (2) why do you start by squaring both sides? What is the rule that you are using to determine that this is the first step?

Re: [tricky one] Is x > 0? [#permalink]
06 Nov 2013, 08:43

Bunuel wrote:

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

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.

Sneaky problem. That's interesting that you're expected to catch this before solving the equation.

I solved for x=2 and x=0 in A and then tested to find that 0 wasn't actually valid (repeating the same process in B to find that A was correct answer).

Also, could you explain again why you will sometimes solve for answers in these absolute value equations that aren't actually valid solutions? That doesn't make sense to me.

Re: [tricky one] Is x > 0? [#permalink]
07 Nov 2013, 04:56

Expert's post

vandygrad11 wrote:

Bunuel wrote:

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

(2) \(|x-3|=|2x-3|\). Square both sides: \((x-3)^2=(2x-3)^2\) --> \((2x-3)^2-(x-3)^2=0\). Apply \(a^2-b^2=(a-b)(a+b)\), rather than squaring: --> \(x(3x-6)=0\) --> \(x=0\) or \(x=2\). Not sufficient.

Answer: A.

Hope it helps.

Sneaky problem. That's interesting that you're expected to catch this before solving the equation.

I solved for x=2 and x=0 in A and then tested to find that 0 wasn't actually valid (repeating the same process in B to find that A was correct answer).

Also, could you explain again why you will sometimes solve for answers in these absolute value equations that aren't actually valid solutions? That doesn't make sense to me.

It depends on how you solve. So, please show your work. _________________

a. |x+3| = 4x -3 |x+3| = x + 3 for x + 3 >= 0 ---> x > -3. Solving for x, x + 3 = 4x -3 ---> 6 = 3x ---> x = 2 ( solution accepted since x > -3)

| x + 3| = -(x +3 ) for x + 3 < 0 --> x < -3. Solving for x, -x - 3 = 4x -3 5x = 0 ---> x = 0 ( solution discarded as x is not < -3)

A is sufficient.

b. |x -3| = |2x -3| |x -3| = x -3 for x-3 >= 0 ---> x >= 3 |x -3| = -(x -3) for x-3 < 0 ---> x < 3

|2x -3 | = 2x -3 for 2x -3 >= 0 ----> x >=3/2 |2x -3 | = -(2x -3) for 2x -3 < 0 ----> x < 3/2

So we have 3 ranges

1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard 2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution ) 3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)

Let me know if I have solved the question correctly. I know the process is lengthy.

One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case 3/2 < x < 3), then both statements are sufficient independently and we get the answer as option (D). _________________

a. |x+3| = 4x -3 |x+3| = x + 3 for x + 3 >= 0 ---> x > -3. Solving for x, x + 3 = 4x -3 ---> 6 = 3x ---> x = 2 ( solution accepted since x > -3)

| x + 3| = -(x +3 ) for x + 3 < 0 --> x < -3. Solving for x, -x - 3 = 4x -3 5x = 0 ---> x = 0 ( solution discarded as x is not < -3)

A is sufficient.

b. |x -3| = |2x -3| |x -3| = x -3 for x-3 >= 0 ---> x >= 3 |x -3| = -(x -3) for x-3 < 0 ---> x < 3

|2x -3 | = 2x -3 for 2x -3 >= 0 ----> x >=3/2 |2x -3 | = -(2x -3) for 2x -3 < 0 ----> x < 3/2

So we have 3 ranges

1) x >=3 ==> x - 3 = 2x -3 ====> x = 0 ; discard 2) 3/2 < x < 3 ===> -(x-3) = 2x -3 ====> -x + 3 = 2x -3 ===> x =2 ( accepted solution ) 3) x < 3/2 ===> -(x-3) = -(2x -3) ====> x = 3 (discarded the solution)

Let me know if I have solved the question correctly. I know the process is lengthy.

One more question, if we consider the only acceptable solution in statement b i.e., x =2 (case 3/2 < x < 3), then both statements are sufficient independently and we get the answer as option (D). tnx for this _________________

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
12 May 2015, 10:56

vikram4689 wrote:

dvinoth86 wrote:

Is x > 0?

(1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|

Liked the question? encourage by giving kudos

Remember: When you have || on both sides of eqn you do not need to VERIFY the answer by putting them back in eqn BUT when you have || on only one side you MUST VERIFY the answer by putting them back in eqn

Re: Is x > 0? (1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3| [#permalink]
12 May 2015, 22:57

1

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

ranaazad wrote:

vikram4689 wrote:

dvinoth86 wrote:

Is x > 0?

(1) |x + 3| = 4x – 3 (2) |x – 3| = |2x – 3|

Liked the question? encourage by giving kudos

Remember: When you have || on both sides of eqn you do not need to VERIFY the answer by putting them back in eqn BUT when you have || on only one side you MUST VERIFY the answer by putting them back in eqn

MBA Acceptance Rate by Country Most top American business schools brag about how internationally diverse they are. Although American business schools try to make sure they have students from...

McCombs Acceptance Rate Analysis McCombs School of Business is a top MBA program and part of University of Texas Austin. The full-time program is small; the class of 2017...