If y=|x+1|/x and x!=0, is xy0?

Question

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

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

(2)  y
eq{0}

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

BrushMyQuant · Accepted Answer

Question can be written as |x+1| = xy 
so, to prove that xy> 0 we essentially need to prove that |x|1| > 0

STAT1  
x^2 + 2x + 1 > 0
means,
(x+1) ^ 2 >0
taking square root on both the sides does not change the inequality
so, we have
|x+1| > 0
which means that xy > 0
So,  SUFFICIENT

STAT2  
y != 0
we know that 
|x+1| = xy => xy is non negative. so, it is either 0 or positive
since we know that both x and y are not equal to zero so, xy > 0
So,  SUFFICIENT

So,  Answer will be D 
Hope it helps!

Watch the following video to learn Basics of Inequalities

https://www.youtube.com/watch?v=1THrnWJtOqc

Check out these posts to understand Absolute Values and Inequalities in Detail

How to Solve: Inequalities (Basic) 
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Narenn · Answer

@ Zarrolou , 
 Regret for not responding within time.

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

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

B) y
eq{0}

First lets simplify the inequality

\frac{|x+1|}{x}  –y = 0 ---------->  \frac{|x+1|-xy}{x}  = 0 -----> we know  x
eq{}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
eq{-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
eq{-1}  and xy>0 ---------------->  Sufficient

Statement 2)  y
eq{0}

From the question stem we know  x
eq{}0 
As per Statement 2,  y
eq{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

nikhilsehgal · Answer

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.....

WholeLottaLove · Answer

Agreed

I made the mistake of multiplying y by x and having to test four different cases:

yx = x+1
-yx = x+1

Then I had to test cases for X>0 and X<0 (i.e. four total cases) and each one turned out to hold true. Your way makes more sense and is quicker to boot.

for 2.) xy = |x+1| so xy could be 0 or greater than zero. We know that X isn't zero from the stem and we know that Y isn't zero from #2. Therefore x/y must be greater or less than zero yielding a positive or negative result. However, because xy = the absolute value of x+1 xy can only be greater than or equil to zero and because both x and y are not zero, product zy MUST be greater than zero.

Good work!

arpanpatnaik · Answer

The answer sure does appear to be

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

Zarrolou · Answer

Good job WholeLottaLove, nktdotgupta, arpanpatnaik !

Official explanation

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

Rewrite the question as (multiply by x)  xy=|x+1|

The  |abs|  is  \geq{0} , so basically we have to check if  xy=|x+1|
eq{0}

A) x^2+2x+1>0  
 (x+1)^2>0 
So  x
eq{-1}  and  |x+1|>0

This is what we are looking for. Sufficient

B) y
eq{0}

y=\frac{|x+1|}{x}
eq{0}  so  |x+1|
eq{0}

Sufficient

The correct answer is D

Hope everything is clear

jlgdr · Answer

Lets make things simpler here.

If y=|x+1|/x, Is xy>0?

Now since |x+1| is always non negative then, xy>0 always except when x+1=0---> x=-1.

Thus, we need to prove that x is not equal to -1.

Statement 1.

We have that (x+1)^2>0

Therefore, we have that |x+1|>0

x+1>0---> x>-1 or x+1<0, x<-1. Therefore, in both cases x is different from -1.

Sufficient

Statement 2

If y is different from zero it means that x is also different from -1, since only the numerator can give 0. Remember x can't be zero in this case.

Sufficient

Answer: D

Hope this clarifies
Cheers!
J

umg · Answer

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
eq{}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
eq{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
eq{0} . 
Therefore,  x
eq{-1}  and as given  y
eq{0}

Hence, xy > 0 (Always)

Statement 2: Sufficient 
 Answer: B

Bunuel · Answer

Notice that x^2+2x+1>0 is true for all value of x except when x = -1. So, for (1)  x
eq -1  and if  x
eq -1 , then  y
eq 0 . Your third case is not possible.