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

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Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]
1
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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]
3
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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]
5
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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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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]
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
Cheers!
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Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]
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]
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]
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]
xy=|x+1|
Is |x+1|>0?
|x+1|=0 for x=-1, and >0 otherwise

S1) (x+1)^2>0 means x>-1, => xy>0(suff)

S2) y=/= 0, means x is not -1, hence |x+1|>0, =>xy>0 (suff)
Re: If y=|x+1|/x and x!=0, is xy>0? [#permalink]
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