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# If y=|x+1|/x and x!=0, is xy>0?

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If y=|x+1|/x and x!=0, is xy>0? [#permalink]

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Updated on: 12 Jul 2014, 12:43
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40% (01:13) correct 60% (01:35) wrong based on 262 sessions

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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

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Originally posted by Zarrolou on 16 May 2013, 11:24.
Last edited by Bunuel on 12 Jul 2014, 12:43, edited 3 times in total.
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Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

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16 May 2013, 11:47
8
2
Zarrolou wrote:
If $$y=\frac{|x+1|}{x}$$ and $$x\neq{}0$$, is $$xy>0$$ ?

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

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

My own question, as always any feedback is appreciated
Kudos to the first correct solution(s)!

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 ar enot equal to zero so, xy > 0
so, SUFFICIENT

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Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

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Updated on: 16 May 2013, 11:59
1
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.....

Originally posted by nikhilsehgal on 16 May 2013, 11:41.
Last edited by nikhilsehgal on 16 May 2013, 11:59, edited 1 time in total.
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Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

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16 May 2013, 12:23
2
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!

nktdotgupta wrote:
Zarrolou wrote:
If $$y=\frac{|x+1|}{x}$$ and $$x\neq{}0$$, is $$xy>0$$ ?

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

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

My own question, as always any feedback is appreciated
Kudos to the first correct solution(s)!

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 ar enot equal to zero so, xy > 0
so, SUFFICIENT

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Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

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16 May 2013, 13:23
3
Zarrolou wrote:
If $$y=\frac{|x+1|}{x}$$ and $$x\neq{}0$$, is $$xy>0$$ ?

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

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

My own question, as always any feedback is appreciated
Kudos to the first correct solution(s)!

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
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Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

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16 May 2013, 13:46
2
1
Good job WholeLottaLove, nktdotgupta, arpanpatnaik !

Official explanation

If $$y=\frac{|x+1|}{x}$$ and $$x\neq{}0$$, 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 $$xy=|x+1|\neq{0}$$

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

This is what we are looking for. Sufficient

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

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

Sufficient

Hope everything is clear
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Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

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16 May 2013, 15:41
1
1
@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

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

Regards,

Narenn

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Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

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24 May 2014, 11:22
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

Hope this clarifies
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Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

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08 Sep 2016, 03:56
BrushMyQuant, arpanpatnaik, Zarrolou, Narenn & jlgdr

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
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Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

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08 Sep 2016, 04:11
umg wrote:
BrushMyQuant, arpanpatnaik, Zarrolou, Narenn & jlgdr

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

Notice that x^2+2x+1>0 is true for all value of x except when x = -1. So, for (1) $$x\neq -1$$ and if $$x\neq -1$$, then $$y\neq 0$$. Your third case is not possible.
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If y=|x+1|/x and x!=0, is xy>0? [#permalink]

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08 Sep 2016, 04:51
Bunuel wrote:
umg wrote:
BrushMyQuant, arpanpatnaik, Zarrolou, Narenn & jlgdr

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

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

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.
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Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]

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17 Sep 2017, 20:58
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Re: If y=|x+1|/x and x!=0, is xy>0?   [#permalink] 17 Sep 2017, 20:58
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