If \(y=\frac{|x+1|}{x}\) and \(x\neq{}0\), is \(xy>0\) ?

(1) \(x^2+2x+1>0\)

(2) \(y\neq{0}\)

My own question, as always any feedback is appreciated
Click here for the OE
_________________

It is beyond a doubt that all our knowledge that begins with experience.

Tips and tricks: Inequalities , Mixture | Review: MGMAT workshop
Strategy: SmartGMAT v1.0 | Questions: Verbal challenge SC I-II- CR New SC set out !! , My Quant

Rules for Posting in the Verbal Forum - Rules for Posting in the Quant Forum[/size][/color][/b]

E it is.....................
For getting xy>0 we need to prove that both x and y are either greater than 0 I.e positive or both of them are negative......

Xy = |x+1|

Once negative:
Xy= -x-1
Xy+x=-1
X=-1/y+1

Once positive: similarly we will get x= 1/y-1

We are getting x once positive and negative ......so can't say xy >0 or not....

II
X^2+2x+1>0
Gives us (x+1)^2>0
Gives us x>-1 and from question stem we know that x is not equal to 0 - hence x will be positive.
But we still don't know their value/ sign of y. Hence this statement is also not sufficient.

Hope I'm correct.....

The answer sure does appear to be [D]

Statement 1: \(x^2+2x+1>0\)
Hence \((x+1)^2\) > 0

Now since the above is a squared term, the value will always be positive except for x = -1. Since the statement doesnt include 0 under the range we can assume that x is not equal to -1. Putting the same values into our main equation i.e. xy = |x+1| we, can be sure that x+1 is not equal to zero at any point. Hence the statement would be sufficient, since if we prove |x+1| is not equal to zero, then its always greater than 0! Hence xy = |x+1| > 0

Statement 2. Now if y is not equal to zero, we can assume |x+1| is never equal to zero! Hence we prove the fact by the same above method, that xy = |x+1| > 0

The idea in the above question, is to establish that at no point x = -1. At only that value y = 0 and xy = 0. Otherwise for all other possible values |x+1| > 0.

As always wonderful question Zarrolou! Hope my procedure is correct!

Regards,
Arpan
_________________

Feed me some KUDOS! *always hungry*

My Thread : Recommendation Letters

@Zarrolou,
Regret for not responding within time.


If \(y=\frac{|x+1|}{x}\) and \(x\neq{}0\), is \(xy>0\) ?

A)\(x^2+2x+1>0\)

B)\(y\neq{0}\)


First lets simplify the inequality

\(\frac{|x+1|}{x}\) –y = 0 ----------> \(\frac{|x+1|-xy}{x}\) = 0 -----> we know \(x\neq{}0\), then |x+1|-xy must be zero. Hence |x+1| - xy = 0 --------> |x+1| = xy


We are asked whether xy>0 --------> Whether |x+1| > 0 ? --------> We know the expression within modules can either be zero or greater than zero. For xy to be greater than zero |x+1| has to be greater than zero.
|x+1| will be zero only when x=-1 and for any other value of x, |x+1| will always be greater than zero
So the question can be rephrased as whether \(x\neq{-1}\)


Statement 1) \(x^2\) + 2x + 1 > 0

This is an quadratic inequality.
Rule :- For any quadratic inequation \(ax^2\) + bx + c > 0, if \(b^2\) – 4ac = 0 and a > 0 then the inequality holds true outside the interval of roots
In our case \(b^2\) – 4ac = 4 – 4 = 0 and a > 0 so \(x^2\) + 2X + 1 > 0 will hold true for all values beyond the Root(s) of equation (Towards any direction - Positive or Negative)
\(x^2\) + 2x + 1 = 0 --------> x(x+1) +1(x + 1) = 0 ---------> (x+1)(x+1) = 0 ----------> x=Root = -1
So \(x^2\) + 2x + 1 > 0 will hold true for any of x except for -1
This reveals that \(x\neq{-1}\) and xy>0 ----------------> Sufficient


Statement 2) \(y\neq{0}\)

From the question stem we know \(x\neq{}0\)
As per Statement 2, \(y\neq{0}\)-------------->That means both X and Y are nonzero.

|x+1| = xy
xy can be either Positive or Negative
|x+1| can be Zero or Positive
Combining both these inferences we can conclude that XY must be Positive. Sufficient

Answer = D

Regards,

Narenn
_________________

Starts from Feb 4th: MBA Video Series, Video answers to specific components and questions about the MBA application.

BrushMyQuant, arpanpatnaik, Zarrolou, Narenn & jlgdr

Are you guys sure that Answer is D? because your own Solutions are Contradicting your answers.

Question: Is xy > 0
Given: \(x\neq{}0\)
\(y=\frac{|x+1|}{x}\)

|x+1| will always be Non-Negative but it can STILL BE 0.
Similarly, y CAN ALSO BE 0.

Statement 1: \(x^2+2x+1>0\)
=> \((x+1)^2 > 0\)
=> x > -1

There is no constraint on y.
So,
Case 1: If y = 1, x > 0;xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 2: y = -1, x = -0.5; xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 3: y = 0, x > -1; xy = 0 i.e. xy !> 0

Statement 1: Insufficient

Statement 2: \(y\neq{0}\)

Case 1: If y = 1, x > 0;xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 2: y = -1, x = -0.5; xy > 0 (x & y must have same sign because |x+1| will always be Non-Negative.)
Case 3: y = -100, x = -1; xy = |x+1| = 0 i.e. y = 0 (because if x = -1 and xy = 0, y = 0). But this case violates the condition given in Statement 2 that \(y\neq{0}\).
Therefore, \(x\neq{-1}\) and as given \(y\neq{0}\)

Hence, xy > 0 (Always)

Statement 2: Sufficient
Answer: B
_________________

I'd appreciate learning about the grammatical errors in my posts

Please hit Kudos If my Solution helps

My Debrief for 750 - https://gmatclub.com/forum/from-720-to-750-one-of-the-most-difficult-pleatues-to-overcome-246420.html

My CR notes - https://gmatclub.com/forum/patterns-in-cr-questions-243450.html

Wow! This question is crazier than I anticipated.

I see it now. I used that logic in Statement 2 but not in Statement 1.

Thanks for pointing it out.
_________________

I'd appreciate learning about the grammatical errors in my posts

Please hit Kudos If my Solution helps

My Debrief for 750 - https://gmatclub.com/forum/from-720-to-750-one-of-the-most-difficult-pleatues-to-overcome-246420.html

My CR notes - https://gmatclub.com/forum/patterns-in-cr-questions-243450.html