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# Is x > 0 ? (1) xy + y = y (2) xy + x = x

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Is x > 0 ? (1) xy + y = y (2) xy + x = x  [#permalink]

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15 Dec 2015, 08:00
1
3
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Difficulty:

45% (medium)

Question Stats:

57% (01:16) correct 43% (01:14) wrong based on 98 sessions

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Is $$x > 0$$ ?

(1) $$xy + y = y$$

(2) $$xy + x = x$$
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Re: Is x > 0 ? (1) xy + y = y (2) xy + x = x  [#permalink]

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15 Dec 2015, 09:09
(1) xy+y=y
y(x+1)=y
x+1=1, so x=0...Suff
(2) xy+x=x
x(y+1)=x
y+1=1, so y=0...then the above eq becomes x(1)=x...here can be a negative, zero or postive....Insuff

Ans: A
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Re: Is x > 0 ? (1) xy + y = y (2) xy + x = x  [#permalink]

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15 Dec 2015, 09:38
1
asethi wrote:
(1) xy+y=y
y(x+1)=y
x+1=1, so x=0...Suff
(2) xy+x=x
x(y+1)=x
y+1=1, so y=0...then the above eq becomes x(1)=x...here can be a negative, zero or postive....Insuff

Ans: A

Hi,
the next step after the coloured portion will not be the one you have written..

1) xy+y=y..
or xy+y-y=0 ...
xy=0...
three scenarios..
a) x=0, y can be any int/fraction..
b)x=0,y=0....
c) x=some int/fraction, y=0...
so nothing can be said about x...insuff..

2) xy+x=x..
or xy+x-x=0..
or xy=0..
same as 1).. insuff

ans E

asethi, do not cancel out terms on either side of equal sign...
bring all terms on one side and then solve for the values...
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Is x > 0 ? (1) xy + y = y (2) xy + x = x  [#permalink]

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03 Mar 2017, 17:58
Is x > 0 ?

(1) xy + y = y

(2) xy + x = x

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
EITHER statement BY ITSELF is sufficient to answer the question.
Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.

In this question, for statement 1, cant we take y common, y(x+1)=y, and then cancel out y, to get x+1 = 0, and thus x= -1?
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Re: Is x > 0 ? (1) xy + y = y (2) xy + x = x  [#permalink]

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03 Mar 2017, 18:14
1
1
anshuverma88 wrote:
Is x > 0 ?

(1) xy + y = y

(2) xy + x = x

Statement (1) BY ITSELF is sufficient to answer the question, but statement (2) by itself is not.
Statement (2) BY ITSELF is sufficient to answer the question, but statement (1) by itself is not.
Statements (1) and (2) TAKEN TOGETHER are sufficient to answer the question, even though NEITHER statement BY ITSELF is sufficient.
EITHER statement BY ITSELF is sufficient to answer the question.
Statements (1) and (2) TAKEN TOGETHER are NOT sufficient to answer the question, requiring more data pertaining to the problem.

In this question, for statement 1, cant we take y common, y(x+1)=y, and then cancel out y, to get x+1 = 0, and thus x= -1?

Dear anshuverma88,

I'm happy to respond.

My friend, I see that you are relatively new to GMAT Club. A few guidelines.

1) The most important is as follows. DO NOT start a new thread for a question that has already been posted. Always search first. Most of the questions from most major test prep companies have already been posted. This question has been posted before here:
https://gmatclub.com/forum/is-x-0-1-xy- ... 10051.html
I will have to ask the genius Bunuel to merge the posts. Find the pre-existent post: it may be that you can find an answer to your question there, and if you can't, that's the appropriate place to post it. Be very reluctant to start a brand new post for a question, and do so only if you can find absolutely no evidence of the question anywhere else on GMAT Club.

2) When you post GMAT DS, there is absolutely no need to post the answer choices. They are always the same, and you should memorize them. See:
GMAT Data Sufficiency Tips

Now, to answer your question. This is a very common algebra mistake. You NEVER can divide by a variable unless you are guaranteed that the variable cannot equal zero. If the variable has a possibility of equaling zero, the dividing by it produces mathematical nonsense. That's what happened to you. See:
GMAT Math: Can you divide by a variable?

If y ≠ 0, then x = -1, but if y = 0, then x could equal anything, including the possibility of zero. That's why the statement is not sufficient.

Does all this make sense?
Mike
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Re: Is x > 0 ? (1) xy + y = y (2) xy + x = x  [#permalink]

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03 Mar 2017, 20:08
anshuverma88 wrote:
Is x > 0 ?

(1) xy + y = y

(2) xy + x = x

In this question, for statement 1, cant we take y common, y(x+1)=y, and then cancel out y, to get x+1 = 0, and thus x= -1?

(1) $$xy+y=y \implies xy=0 \implies x=0$$ or $$y=0$$. We can't deduce that whether $$x>0$$ or not. Insufficient.

To clear your confusion, $$xy+y=y \implies y(x+1)=y \; (*)$$.

From (*), we can't clear $$y$$ to have $$x+1=0$$. The correct way is:

If $$y=0$$, then $$x \in R$$. This means $$x$$ could be any value.
If $$y \neq 0$$, then we have $$x+1 = 1$$, not $$0$$ as you said.

The better way is:
$$y(x+1)=y \implies y(x+1)-y=0 \implies y\times x =0$$.

(2) $$xy + x = x \implies xy=0$$, the same as (1). Insufficient.

Combine (1) and (2), we still have $$xy=0$$. Insufficient.

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Re: Is x > 0 ? (1) xy + y = y (2) xy + x = x  [#permalink]

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04 Mar 2017, 05:05
Statement analysis
St 1: xy +y = y or xy = 0 . that means either x is 0 or y is zero. INSUFFICIENT
St 2: xy +x = x or xy = 0. that means either x is 0 or y is zero. INSUFFICIENT

St 1 & St 2: both are yielding xy = 0, it is again insufficient.
Option E
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Re: Is x > 0 ? (1) xy + y = y (2) xy + x = x  [#permalink]

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10 Apr 2019, 20:55
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Re: Is x > 0 ? (1) xy + y = y (2) xy + x = x   [#permalink] 10 Apr 2019, 20:55
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