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If |2x + 5| = |3x − 2|, which of the following is a possible value of

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If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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If |2x + 5| = |3x − 2|, which of the following is a possible value of x?

(a) -7
(b) -19/5
(c) -3/5
(d) 3/5
(e) 5

Source: Veritas Prep Mock
[Reveal] Spoiler: OA

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Last edited by Bunuel on 21 Nov 2017, 00:24, edited 1 time in total.
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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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New post 19 Nov 2017, 17:10
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adkikani wrote:
If |2x+5|=|3x−2|, which of the following is a possible value of x?

(a) -7
(b) -195
(c) -35
(d) 35
(e) 5

Source: Veritas Prep Mock


Did you transcribe the question (or answer choices) correctly?

Cheers,
Brent
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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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New post 19 Nov 2017, 17:15
GMATPrepNow

Apologize for the same and corrected.
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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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New post 19 Nov 2017, 17:35
adkikani wrote:
If |2x+5|=|3x−2|, which of the following is a possible value of x?

(a) -7
(b) -19/5
(c) -3/5
(d) 3/5
(e) 5

Source: Veritas Prep Mock


2x+5 = 3x-2
x= 7
2x+5 = -3x+2
x=-3/5

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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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New post 19 Nov 2017, 17:47
adkikani wrote:
If |2x+5|=|3x−2|, which of the following is a possible value of x?

(a) -7
(b) -19/5
(c) -3/5
(d) 3/5
(e) 5

Source: Veritas Prep Mock



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

C
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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New post 19 Nov 2017, 17:48
adkikani wrote:
If |2x+5|=|3x−2|, which of the following is a possible value of x?

(a) -7
(b) -19/5
(c) -3/5
(d) 3/5
(e) 5

Source: Veritas Prep Mock

After plugging in Answers A and E because the math was quick, and finding neither to be a solution, I took the case approach.
+LHS = +RHS
+LHS = -RHS

|2x+5|=|3x−2|

Case 1:
2x + 5 = 3x - 2
7 = x

Case 2:
2x + 5 = -(3x-2)
2x + 5 = -3x + 2
5x = -3
x = -3/5

Answer C

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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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New post 19 Nov 2017, 22:38
adkikani wrote:
If |2x+5|=|3x−2|, which of the following is a possible value of x?

(a) -7
(b) -19/5
(c) -3/5
(d) 3/5
(e) 5

Source: Veritas Prep Mock


Hi adkikani

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.

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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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New post 19 Nov 2017, 23:12
niks18

Thanks for your explanation. However I really found it a bit intimidating. ;)

Is rahul16singh28 solution technically correct too?

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

Let me know your inputs :-)
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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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New post 20 Nov 2017, 00:11
adkikani wrote:
niks18

Thanks for your explanation. However I really found it a bit intimidating. ;)

Is rahul16singh28 solution technically correct too?

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

Let me know your inputs :-)


Hi adkikani,

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 ;) :-)

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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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New post 21 Nov 2017, 00:20
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adkikani wrote:
If |2x+5|=|3x−2|, which of the following is a possible value of x?

(a) -7
(b) -19/5
(c) -3/5
(d) 3/5
(e) 5

Source: Veritas Prep Mock


Method 1:

If I get this question in the test, I would just substitute options in the equation. That's the fastest way.

|2x+5|=|3x−2|
Try the integer values very quickly first. They don't satisfy

|2x + 25/5| = |3x - 10/5|
Now try 3/5, then -3/5.

-3/5 satisfies. Answer (C)


Method 2: |2x+5|=|3x−2|

2 * |x + 5/2| = 3 * |x - 2/3|

____________________ ( - 15/6) _________________________ ( 0 ) ____________ ( 4/6 )_____________________

This is a distance of 19 between them. Twice the distance from -15/6 should be equal to thrice the distance from 4/6.

So the point will be \(\frac{-15}{6} + (\frac{19}{6}*\frac{3}{5})\)

We get \(\frac{-3}{5}\)
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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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New post 21 Nov 2017, 00:34
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Quote:
Method 1:

If I get this question in the test, I would just substitute options in the equation. That's the fastest way.

|2x+5|=|3x−2|
Try the integer values very quickly first. They don't satisfy

|2x + 25/5| = |3x - 10/5|
Now try 3/5, then -3/5.

-3/5 satisfies. Answer (C)


I believe you also sensed that having 5 in denominator will help you ease the addition/ subtraction of fractions.
Super quick, thanks a ton. :thumbup:

Quote:
Method 2: |2x+5|=|3x−2|

2 * |x + 5/2| = 3 * |x - 2/3|

____________________ ( - 15/6) _________________________ ( 0 ) ____________ ( 4/6 )_____________________

This is a distance of 19 between them. Twice the distance from -15/6 should be equal to thrice the distance from 4/6.

So the point will be \(\frac{-15}{6} + (\frac{19}{6}*\frac{3}{5})\)

We get \(\frac{-3}{5}\)


I totally could not understand method 2. Can you elaborate?
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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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New post 21 Nov 2017, 08:35
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adkikani wrote:
VeritasPrepKarishma



Quote:
Method 1:

If I get this question in the test, I would just substitute options in the equation. That's the fastest way.

|2x+5|=|3x−2|
Try the integer values very quickly first. They don't satisfy

|2x + 25/5| = |3x - 10/5|
Now try 3/5, then -3/5.

-3/5 satisfies. Answer (C)


I believe you also sensed that having 5 in denominator will help you ease the addition/ subtraction of fractions.
Super quick, thanks a ton. :thumbup:

Quote:
Method 2: |2x+5|=|3x−2|

2 * |x + 5/2| = 3 * |x - 2/3|

____________________ ( - 15/6) _________________________ ( 0 ) ____________ ( 4/6 )_____________________

This is a distance of 19 between them. Twice the distance from -15/6 should be equal to thrice the distance from 4/6.

So the point will be \(\frac{-15}{6} + (\frac{19}{6}*\frac{3}{5})\)

We get \(\frac{-3}{5}\)


I totally could not understand method 2. Can you elaborate?


This post explains you what absolute value is:
http://www.veritasprep.com/blog/2011/01 ... edore-did/
And this one discusses how to use this concept:
https://www.veritasprep.com/blog/2011/0 ... s-part-ii/

In this question, we say that x is a point such that twice its distance from -5/2 is equal to thrice its distance from 2/3.

So x will divide the distance between the two points in the ratio 3:2. Hence, x will be at a distance 3/5th to the right from -15/6.

Does this help?
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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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New post 22 Nov 2017, 11:18
adkikani wrote:
If |2x + 5| = |3x − 2|, which of the following is a possible value of x?

(a) -7
(b) -19/5
(c) -3/5
(d) 3/5
(e) 5


First we can solve for when both 2x+5 and 3x−2 are positive:

|2x + 5| = |3x − 2|

2x + 5 = 3x - 2

-x = -7

x = 7

Next we can solve for when 2x+5 is positive and 3x−2 is negative.

|2x + 5| = |3x − 2|

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

2x + 5 = -3x + 2

5x = -3

x = -3/5

Answer: C
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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of [#permalink]

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New post 23 Nov 2017, 09:23
niks18 wrote:
adkikani wrote:
niks18

Thanks for your explanation. However I really found it a bit intimidating. ;)

Is rahul16singh28 solution technically correct too?

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

Let me know your inputs :-)


Hi adkikani,

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.

Cheers,
Kabir

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Re: If |2x + 5| = |3x − 2|, which of the following is a possible value of   [#permalink] 23 Nov 2017, 09:23
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