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If 6xy = x^2y + 9y, what is the value of xy?

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If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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Updated on: 16 Jul 2015, 00:55
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If 6xy = $$x^2$$y + 9y, what is the value of xy?

(1) x = –2
(2) x < 0

The answer is D, with the following explanation:
The equation in question can be rephrased as follows:

x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0

Therefore, one or both of the following must be true:
y = 0 or
x = 3

It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"

(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0

Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.

(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...

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Originally posted by MisterEko on 19 Dec 2010, 16:04.
Last edited by reto on 16 Jul 2015, 00:55, edited 1 time in total.
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Re: If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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19 Dec 2010, 16:22
13
3
MisterEko wrote:
Guys, I just have a question if my way of solving it is reasonable and can often be applied here. I got the right answer on this one, but am not sure it would work usually:

If 6xy = x2y + 9y, what is the value of xy?

(1) x = –2

(2) x < 0

The answer is D, with the following explanation:
The equation in question can be rephrased as follows:

x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0

Therefore, one or both of the following must be true:
y = 0 or
x = 3

It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"

(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0

Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.

(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...

For statement (1) your reasoning looks OK. As for statement (2) what do you mean by "did the same"?

One can also do this way: $$6xy=x^2y + 9y$$ --> $$y(x^2-6x+9)=0$$ --> $$y(x-3)^2=0$$ --> either $$x=3$$ or $$y=0$$ (or both).

(1) x = –2 --> so $$x\neq{3}$$, then $$y=0$$ and $$xy=0$$. Sufficient.

(2) x < 0 --> again $$x\neq{3}$$, then $$y=0$$ and $$xy=0$$. Sufficient.

Check question #1 here: inequality-and-absolute-value-questions-from-my-collection-86939.html for harder version of the same problem.

Hope it helps.
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Re: If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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19 Dec 2010, 17:01
Bunuel wrote:
MisterEko wrote:
Guys, I just have a question if my way of solving it is reasonable and can often be applied here. I got the right answer on this one, but am not sure it would work usually:

If 6xy = x2y + 9y, what is the value of xy?

(1) x = –2

(2) x < 0

The answer is D, with the following explanation:
The equation in question can be rephrased as follows:

x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0

Therefore, one or both of the following must be true:
y = 0 or
x = 3

It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"

(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0

Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.

(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...

For statement (1) your reasoning looks OK. As for statement (2) what do you mean by "did the same"?

One can also do this way: $$6xy=x^2y + 9y$$ --> $$y(x^2-6x+9)=0$$ --> $$y(x-3)^2=0$$ --> either $$x=3$$ or $$y=0$$ (or both).

(1) x = –2 --> so $$x\neq{3}$$, then $$y=0$$ and $$xy=0$$. Sufficient.

(2) x < 0 --> again $$x\neq{3}$$, then $$y=0$$ and $$xy=0$$. Sufficient.

Check question #1 here: inequality-and-absolute-value-questions-from-my-collection-86939.html for harder version of the same problem.

Hope it helps.

Thanks a lot. What I meant by "I did the same" is that I substituted x for -1 and tested, and result came out to be the same...
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http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142

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Re: If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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19 Dec 2010, 23:53
MisterEko wrote:
Thanks a lot. What I meant by "I did the same" is that I substituted x for -1 and tested, and result came out to be the same...

Yes, for ANY x but 3 you'll get ky=0 for some nonzero k and thus y=0 (for x=3 you'll get 0=0).
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Re: If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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11 Dec 2013, 03:14
1
Bunuel wrote:
MisterEko wrote:
Guys, I just have a question if my way of solving it is reasonable and can often be applied here. I got the right answer on this one, but am not sure it would work usually:

If 6xy = x2y + 9y, what is the value of xy?

(1) x = –2

(2) x < 0

The answer is D, with the following explanation:
The equation in question can be rephrased as follows:

x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0
Therefore, one or both of the following must be true:
y = 0 or
x = 3

It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"

(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0

Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.

(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...

For statement (1) your reasoning looks OK. As for statement (2) what do you mean by "did the same"?

One can also do this way: $$6xy=x^2y + 9y$$ --> $$y(x^2-6x+9)=0$$ --> $$y(x-3)^2=0$$ --> either $$x=3$$ or $$y=0$$ (or both).

(1) x = –2 --> so $$x\neq{3}$$, then $$y=0$$ and $$xy=0$$. Sufficient.

(2) x < 0 --> again $$x\neq{3}$$, then $$y=0$$ and $$xy=0$$. Sufficient.

Check question #1 here: inequality-and-absolute-value-questions-from-my-collection-86939.html for harder version of the same problem.

Hope it helps.

Hi Bunuel,

I am still not clear why do we need statements. Once we know that y=0 then no matter what the value of x is xy will always be =0. Do we need to even consider the statements. Am I thinking in the right direction or is there a disconnect somewhere.
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Posts: 50627
Re: If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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11 Dec 2013, 03:31
2
davidfrank wrote:
Bunuel wrote:
MisterEko wrote:
Guys, I just have a question if my way of solving it is reasonable and can often be applied here. I got the right answer on this one, but am not sure it would work usually:

If 6xy = x2y + 9y, what is the value of xy?

(1) x = –2

(2) x < 0

The answer is D, with the following explanation:
The equation in question can be rephrased as follows:

x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0
Therefore, one or both of the following must be true:
y = 0 or
x = 3

It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"

(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0

Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.

(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...

For statement (1) your reasoning looks OK. As for statement (2) what do you mean by "did the same"?

One can also do this way: $$6xy=x^2y + 9y$$ --> $$y(x^2-6x+9)=0$$ --> $$y(x-3)^2=0$$ --> either $$x=3$$ or $$y=0$$ (or both).

(1) x = –2 --> so $$x\neq{3}$$, then $$y=0$$ and $$xy=0$$. Sufficient.

(2) x < 0 --> again $$x\neq{3}$$, then $$y=0$$ and $$xy=0$$. Sufficient.

Check question #1 here: inequality-and-absolute-value-questions-from-my-collection-86939.html for harder version of the same problem.

Hope it helps.

Hi Bunuel,

I am still not clear why do we need statements. Once we know that y=0 then no matter what the value of x is xy will always be =0. Do we need to even consider the statements. Am I thinking in the right direction or is there a disconnect somewhere.

See the red part is not true. We don't know whether $$y=0$$. From $$y(x-3)^2=0$$ we have that $$x=3$$ OR $$y=0$$. Now, if $$x=3$$, then y can be any number, thus xy is not necessarily 0.

Hope it's clear.
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Re: If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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11 Dec 2013, 05:21
1
2
MisterEko wrote:
If 6xy = x^2y + 9y, what is the value of xy?

(1) x = –2
(2) x < 0

Responding to a pm:

Why do we need the statements?

Given: $$6xy = x^2y + 9y$$
Given: $$y(x-3)^2 = 0$$

What we know is this: EITHER y is 0 OR x is 3. It is possible that both y is 0 and x is 3 but at least one of them MUST be true. This is all we know. We don't know whether y is 0 or whether x is 3 or both.

Question: what is xy?
I cannot say yet. All I know is that either y is 0 or x is 3.
If y is 0, xy will be 0.
If x is 3, then I don't know y which could be anything so xy could be anything.
So I cannot say what xy is.

(1) x = –2

This tells me that x is not 3. If x is not 3, y MUST be 0 because one of them has to be true. If y is 0, we know for sure that xy is 0. Sufficient.

(2) x < 0

This again tells me that x is not 3. If x is not 3, y MUST be 0 because one of them has to be true. If y is 0, we know for sure that xy is 0. Sufficient.

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Re: If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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17 Sep 2015, 07:53
Bunuel wrote:
MisterEko wrote:
Guys, I just have a question if my way of solving it is reasonable and can often be applied here. I got the right answer on this one, but am not sure it would work usually:

If 6xy = x2y + 9y, what is the value of xy?

(1) x = –2

(2) x < 0

The answer is D, with the following explanation:
The equation in question can be rephrased as follows:

x2y – 6xy + 9y = 0
y(x2 – 6x + 9) = 0
y(x – 3)2 = 0

Therefore, one or both of the following must be true:
y = 0 or
x = 3

It follows that the product xy must equal either 0 or 3y. This question can therefore be rephrased several different ways: "What is y?" or "Does x = 3?"

(1) SUFFICIENT: If x = -2, then:
(–2)2y – 6(–2)y + 9y = 0
4y + 12y + 9y = 0
25y = 0

Therefore y = xy = 0. Tn fact, for any value of x other than 3, y must equal zero.

(2) SUFFICIENT: x is negative. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

What I did, i simply applied the info from statement 1, and got that -12Y=13Y which can only be true if Y is 0 and therefore, XY is 0.
In the second statement, i did the same and the result came to be the same. Is this faulty reasoning? Also, this is said to be 700-800 level question and my way seems just too easy to solve such a high level problem...

For statement (1) your reasoning looks OK. As for statement (2) what do you mean by "did the same"?

One can also do this way: $$6xy=x^2y + 9y$$ --> $$y(x^2-6x+9)=0$$ --> $$y(x-3)^2=0$$ --> either $$x=3$$ or $$y=0$$ (or both).

(1) x = –2 --> so $$x\neq{3}$$, then $$y=0$$ and $$xy=0$$. Sufficient.

(2) x < 0 --> again $$x\neq{3}$$, then $$y=0$$ and $$xy=0$$. Sufficient.

Check question #1 here: inequality-and-absolute-value-questions-from-my-collection-86939.html for harder version of the same problem.

Hope it helps.

why we can not cancel out Y from 6xy=x^2y + 9y ? this is not inequality..what is mistake in my thinking
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Re: If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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17 Sep 2015, 19:24
4

why we can not cancel out Y from 6xy=x^2y + 9y ? this is not inequality..what is mistake in my thinking

Usually, it is not a good idea to cancel out a variable. You lose a solution if you do.

If you do not cancel out the y, you get that this inequality: 6xy=x^2y + 9y holds when either x = 3 or y = 0.

If you do cancel out the y, you get that this inequality: 6xy=x^2y + 9y holds when x = 3.

You lost the y = 0 value. Hence, it's always better to take out y common and keep it on the side, not cancel it.
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If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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31 Dec 2017, 05:42
1
[quote="MisterEko"]If 6xy = $$x^2$$y + 9y, what is the value of xy?

(1) x = –2
(2) x < 0

[spoiler=]The answer is D, with the following explanation:
The equation in question can be rephrased as follows:

Slightly different analysis. Bunuel / Experts pl. validate

Given $$6xy=x^2y+9y$$

Find xy= ?

Now $$6xy=x^2y+9y$$ ------ (1)
=> $$6xy=y(x^2+9)$$
=> $$6x=x^2+9$$
=> $$x^2-6x+9=0$$
=> $$x^2-3x-3x+9=0$$
=> $$(x-3)^2=0$$
=> Therefore x=3
=> Substituting x=3 in (1) we have
=> 0=0

So we know the value of 'x'. We need to find y=?

Statement 1 x=-2
=> Substituting in (1) we have
=> -12y=13y
=> OR 25y=0
=> OR y=0
=> Therefore xy=0 SUFFICIENT

Statement 2 x<0 i.e 'x' is -ve
=> Substituting value of 'x' in equ (1)
=> Now irrespective of the magnitude ' -x ' LHS of equ (1) will be -ve
=> i.e 6xy= -ky ---- for some constant 'k'
=> Again irrespective of the magnitude ' -x ' RHS of equ (1) will be +ve
=> i.e $$x^2y+9y=my$$ ---- for some constant 'm'
=> -ky=my
=> OR my+ky=0
=> OR ny=0 ----- for some constant 'n'
=> Irrespective of the SIGN of n, y=0
=> Therefore xy=0. SUFFICIENT

Therefore "D"

Thanks
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If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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10 Apr 2018, 11:13
VeritasPrepKarishma Bunuel niks18 amanvermagmat chetan2u

Quote:
Usually, it is not a good idea to cancel out a variable. You lose a solution if you do.

If you do not cancel out the y, you get that this inequality: 6xy=x^2y + 9y holds when either x = 3 or y = 0.

Do we not fundamentally need another set of quadratic equations since two unknown variables are present.

Eg. $$x^2$$ - 1 = 0

$$x^2$$ = 1 (Adding a positive number does not change inequality)

|x| = 1 or x = 1 / -1 Note: Still here, I am not getting UNIQUE values of x

But the story changes dramatically if $$x^2$$ + y - 1 = 0 ... (2)

How can I get unique solutions for x and y for (2)
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If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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10 Apr 2018, 17:31
VeritasPrepKarishma Bunuel niks18 amanvermagmat

Quote:
Usually, it is not a good idea to cancel out a variable. You lose a solution if you do.

If you do not cancel out the y, you get that this inequality: 6xy=x^2y + 9y holds when either x = 3 or y = 0.

Do we not fundamentally need another set of quadratic equations since two unknown variables are present.

Eg. $$x^2$$ - 1 = 0

$$x^2$$ = 1 (Adding a positive number does not change inequality)

|x| = 1 or x = 1 / -1 Note: Still here, I am not getting UNIQUE values of x

But the story changes dramatically if $$x^2$$ + y - 1 = 0 ... (2)

How can I get unique solutions for x and y for (2
)

I assume that highlighted section does not refer to the question per se but is just an illustration.

There is a difference between your illustration and this question. This question can be factorized to $$y*(x-3)^2=0$$. Now when you get this equation, then there could be only two solutions - either y=0 or x=3. Then from the statements you know the boundary of x which will eventually lead to y=0.

In your illustration $$x^2+y-1=0$$, we know from equation 1 that $$x^2=1$$, hence you can substitute the value of $$x^2$$ in equation (2) to get $$1+y-1=0 => y=0$$
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Re: If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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10 Apr 2018, 18:21
niks18

Thanks for prompt support.

Quote:
I assume that highlighted section does not refer to the question per se but is just an illustration.

Correct, we are on same page.

Quote:
There is a difference between your illustration and this question. This question can be factorized to $$y*(x-3)^2=0$$. Now when you get this equation, then there could be only two solutions - either y=0 or x=3. Then from the statements you know the boundary of x which will eventually lead to y=0.

In your illustration $$x^2+y-1=0$$, we know from equation 1 that $$x^2=1$$, hence you can substitute the value of $$x^2$$ in equation (2) to get $$1+y-1=0 => y=0$$

Apologize to have not mentioned earlier about possibility of (2) being solved independently.
I could, in no way, solve (2) independently was my point.

My two basic queries are :
1. In a typical DS question as in this, where in we are asked to find product of x and y,
I need unique values for both x and y. (Since this is a value Q)

As per VeritasPrepKarishma we DO GET unique values of both x and y.
She did not mention MULTIPLE values of x as in below theoretical example.

$$x^2$$ - 1 = 0

x = 1, -1
So, why do I care if I am loosing solution of another variable?

2. Why is factorization so crucial in simplifying in original q stem?

6xy=$$x^2$$y + 9y

We know numerical values CAN NOT be equal to zero.
So Q stem reduces to :
Is xy = $$x^2$$y +y

The above equation is NOT solvable on its own, not because I am loosing a solution,
but because I can get multiple ways to solve this equality since we are NOT given that
either x or y are integers.

Let me know if my approach is correct.
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Re: If 6xy = x^2y + 9y, what is the value of xy?  [#permalink]

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10 Apr 2018, 18:43
niks18

Thanks for prompt support.

Quote:
I assume that highlighted section does not refer to the question per se but is just an illustration.

Correct, we are on same page.

Quote:
There is a difference between your illustration and this question. This question can be factorized to $$y*(x-3)^2=0$$. Now when you get this equation, then there could be only two solutions - either y=0 or x=3. Then from the statements you know the boundary of x which will eventually lead to y=0.

In your illustration $$x^2+y-1=0$$, we know from equation 1 that $$x^2=1$$, hence you can substitute the value of $$x^2$$ in equation (2) to get $$1+y-1=0 => y=0$$

Apologize to have not mentioned earlier about possibility of (2) being solved independently.
I could, in no way, solve (2) independently was my point.

My two basic queries are :
1. In a typical DS question as in this, where in we are asked to find product of x and y,
I need unique values for both x and y. (Since this is a value Q)

As per VeritasPrepKarishma we DO GET unique values of both x and y.
She did not mention MULTIPLE values of x as in below theoretical example.

$$x^2$$ - 1 = 0

x = 1, -1
So, why do I care if I am loosing solution of another variable?

2. Why is factorization so crucial in simplifying in original q stem?

6xy=$$x^2$$y + 9y

We know numerical values CAN NOT be equal to zero.
So Q stem reduces to :
Is xy = $$x^2$$y +y

The above equation is NOT solvable on its own, not because I am loosing a solution,
but because I can get multiple ways to solve this equality since we are NOT given that
either x or y are integers.

Let me know if my approach is correct.

The highlighted portion is incorrect.

$$6xy=x^2y+9y$$ cannot be simply reduced to $$6xy=x^2y+9y$$. The co-efficient 6 & 9 are an integral part of the equation. In equality you can divide both sides of the equation by same constant to simplify the equation.

also note that in the equation $$6xy=x^2y+9y$$, you have two variables, x & y, so you need to solve for both and that can be done only by factorizing the equation. Here you cannot simply divide both sides of the equation by y to get $$6x=x^2+9$$ because if y=0, then division by 0 is simply not possible. Hence you should never cancel out common "Variables". if you have common "constants/numbers", then you can cancel out them.

As in this question you are required to find the product xy, so you need values of both x & y. Suppose if you only know that x=1 or -1, then xy=y or -y. this will not give you any solution. if y=0, then irrespective of the value of x, xy=0. if you do not care about the variable y, then you will get option E or C as an answer, which in this case will be incorrect.

So, DS question should be looked in TOTALITY and not on the basis of any one of the equation or statement.
Re: If 6xy = x^2y + 9y, what is the value of xy? &nbs [#permalink] 10 Apr 2018, 18:43
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