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Is x > 0?

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Is x > 0?  [#permalink]

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New post Updated on: 01 Aug 2013, 03:43
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Is \(x > 0\)?

(1) \(|x + 3| = 4x - 3\)

(2) \(|x + 1| = 2x - 1\)


Now, I approached the question in the following manner:

Statement 1:
|x+3|>=0 always
So,

4x-3>=0
so X>=3/4

Sufficient

Statement 2

similarly 2X-1>=0
x>1/2
Sufficient

So: D each statement is enough.

Can you tell me whether I am missing something conceptually?
Or a case in which this approach might fail?

Will i be better served using the traditional long approach , solving for X and substituting to see if it satisfies the equation??


Thanks..

Originally posted by 12bhang on 01 Aug 2013, 03:31.
Last edited by Zarrolou on 01 Aug 2013, 03:43, edited 2 times in total.
Edited the question.
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Re: Is x > 0?  [#permalink]

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New post 06 May 2014, 07:52
3
3
jlgdr wrote:
Shortcut LHS, RHS for this one anyone?

Cheers!
J :)


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 smae here: LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(2x-1\geq{0}\) --> \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient.

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

Similar question to practice: is-x-0-1-x-3-4x-3-2-x-3-2x-127978.html
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Senior Manager
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Re: Is x > 0?  [#permalink]

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New post 01 Aug 2013, 03:41
4
12bhang wrote:
Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x + 1| = 2x – 1


Statement 1
If |x + 3| >= 0 -----> x+3 = 4x-3 ----->3x=6----->x=2
If |x + 3| < 0 -----> -x-3 = 4x-3 ----->5x=0----->x=0
x=0 is not possible because we assumed that x+3<0 or x<-3. Thus the only possible value of x is 2.
Sufficient

Statement 2
If |x + 1| >= 0 -----> x+1 = 2x-1 ----->x=2----->x=2
If |x + 1| < 0 -----> -x-1 = 2x-1 ----->3x=0----->x=0
x=0 is not possible because we assumed that x+1<0 or x<-1. Thus the only possible value of x is 2.
Sufficient

Answer D
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Re: Is x > 0?  [#permalink]

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New post 01 Aug 2013, 06:04
2
fameatop wrote:
12bhang wrote:
Is x > 0?
(1) |x + 3| = 4x – 3
(2) |x + 1| = 2x – 1


Statement 1
If |x + 3| >= 0 -----> x+3 = 4x-3 ----->3x=6----->x=2
If |x + 3| < 0 -----> -x-3 = 4x-3 ----->5x=0----->x=0
x=0 is not possible because we assumed that x+3<0 or x<-3. Thus the only possible value of x is 2.
Sufficient

Statement 2
If |x + 1| >= 0 -----> x+1 = 2x-1 ----->x=2----->x=2
If |x + 1| < 0 -----> -x-1 = 2x-1 ----->3x=0----->x=0
x=0 is not possible because we assumed that x+1<0 or x<-1. Thus the only possible value of x is 2.
Sufficient

Answer D



Ok. that is the correct conventional way. Here's what i thought.


|x+3| will always be either > or = to 0.
So , 4x-3>=0
So, x>3/4, which is greater than 0.
So sufficient.

statement 2:
|x+1|>=0
So , 2x-1>=0
or x>=1/2
which is greater than 0
so sufficient

D

What I want to know is whether there is any loophole in my approach. I.e am i missing out on some hidden value of the modulus or something and whether it is better to solve for X and then check whether it falls in the range.

Please help.

Thanks
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Re: Is x > 0?  [#permalink]

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New post 03 Aug 2013, 04:32
12bhang wrote:
Is \(x > 0\)?

(1) \(|x + 3| = 4x - 3\)

(2) \(|x + 1| = 2x - 1\)


Now, I approached the question in the following manner:

Statement 1:
|x+3|>=0 always
So,

4x-3>=0
so X>=3/4

Sufficient

Statement 2

similarly 2X-1>=0
x>1/2
Sufficient

So: D each statement is enough.

Can you tell me whether I am missing something conceptually?
Or a case in which this approach might fail?

Will i be better served using the traditional long approach , solving for X and substituting to see if it satisfies the equation??


Thanks..

this is a wrong process. 3/4 or 1/2 never satisfies the two equations of statements.... use x = 3/4 0r 1/2 in those equations you will see these aren't the solutions.......
some good process already members publishes here, use them.........
thanks
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Re: Is x > 0?  [#permalink]

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New post 04 Aug 2013, 02:53
Asifpirlo wrote:
12bhang wrote:
Is \(x > 0\)?

(1) \(|x + 3| = 4x - 3\)

(2) \(|x + 1| = 2x - 1\)


Now, I approached the question in the following manner:

Statement 1:
|x+3|>=0 always
So,

4x-3>=0
so X>=3/4

Sufficient

Statement 2

similarly 2X-1>=0
x>1/2
Sufficient

So: D each statement is enough.

Can you tell me whether I am missing something conceptually?
Or a case in which this approach might fail?

Will i be better served using the traditional long approach , solving for X and substituting to see if it satisfies the equation??


Thanks..

this is a wrong process. 3/4 or 1/2 never satisfies the two equations of statements.... use x = 3/4 0r 1/2 in those equations you will see these aren't the solutions.......
some good process already members publishes here, use them.........
thanks



Hi, Thanks for your insight.

I agree that the equation does not hold true at the given values. However, the question does not need us to find the actual values.

merely proving that the value of x>0 is enough. This will be proved, as the LHS in both statements are in the modulus and will return values that are greater than or equal to zero.
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Re: Is x > 0?  [#permalink]

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New post 26 Apr 2014, 11:56
Shortcut LHS, RHS for this one anyone?

Cheers!
J :)
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Re: Is x > 0?  [#permalink]

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New post 19 May 2014, 11:32
Bunuel wrote:
jlgdr wrote:
Shortcut LHS, RHS for this one anyone?

Cheers!
J :)


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 smae here: LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(2x-1\geq{0}\) --> \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient.

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

Similar question to practice: is-x-0-1-x-3-4x-3-2-x-3-2x-127978.html


Is it because you have 2x - 1 and therefore 2x>=1, x>=1/2?

What if you had 2x + 1? Will it still work? Don't thimk so right? We would need to analyze both scenarios

Please advice
Cheers!
J :)
Math Expert
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Joined: 02 Sep 2009
Posts: 51218
Re: Is x > 0?  [#permalink]

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New post 20 May 2014, 00:34
1
jlgdr wrote:
Bunuel wrote:
jlgdr wrote:
Shortcut LHS, RHS for this one anyone?

Cheers!
J :)


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 smae here: LHS is an absolute value, which is always non-negative (\(|some \ expression|\geq{0}\)), so RHS must also be non-negative --> \(2x-1\geq{0}\) --> \(x\geq{\frac{1}{2}}\), hence \(x>0\). Sufficient.

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

Similar question to practice: is-x-0-1-x-3-4x-3-2-x-3-2x-127978.html


Is it because you have 2x - 1 and therefore 2x>=1, x>=1/2?

What if you had 2x + 1? Will it still work? Don't thimk so right? We would need to analyze both scenarios

Please advice
Cheers!
J :)


We could still solve \(|x + 1| = 2x + 1\) with this approach to get x. LHS is an absolute value, which is always non-negative, so RHS must also be non-negative --> \(2x+1\geq{0}\) --> \(x\geq{-\frac{1}{2}}\). For, \(x\geq{-\frac{1}{2}}\), \(x+1>0\), hence \(|x + 1| = x + 1\) --> \(x + 1 = 2x + 1\) --> \(x=0\).

But if it were \(|x - 1| = 2x + 1\), then getting \(x\geq{-\frac{1}{2}}\) from LHS, wouldn't be useful to determine whether \(|x - 1| = x - 1\) or \(|x - 1| = -(x - 1)\) because \(x-1\) could be positive as well as negative for different x'es which are more than or equal to -1/2. Therefore we should use usual method of solving for \(|x - 1| = 2x + 1\), not RHS/LHS.

So, this method is not universal and can be applied only when non-modulus part defines x in such way that helps to get the sign of absolute value expression on the other side.

Does this make sense?
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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