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Re: If 6xy = x^2*y + 9y, what is the value of xy? [#permalink]
05 Jun 2013, 04:13

6

This post received KUDOS

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This post was BOOKMARKED

The equation in question can be rephrased as follows: x^2 y – 6xy + 9y = 0

y(x^2 –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 “What is y?”

(1) INSUFFICIENT: This equation cannot be manipulated or combined with the original equation to solve directly for x or y. Instead, plug the two possible scenarios from the original equation into the equation from this statement:

If x = 3, then y = 3 + x = 3 + 3 = 6, so xy = (3)(6) = 18. If y = 0, then x = y – 3 = 0 – 3 = -3, so xy = (-3)(0) = 0. Since there are two possible answers, this statement is not sufficient.

(2) SUFFICIENT: If x3 < 0, then x < 0. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

Ans B

Please someone confirm on the solution, if the approach is correct _________________

Correct me If I'm wrong !! looking for valuable inputs

Re: If 6xy = x^2*y + 9y, what is the value of xy? [#permalink]
05 Jun 2013, 04:25

3

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

Manhnip wrote:

The equation in question can be rephrased as follows: x^2 y – 6xy + 9y = 0

y(x^2 –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 “What is y?”

(1) INSUFFICIENT: This equation cannot be manipulated or combined with the original equation to solve directly for x or y. Instead, plug the two possible scenarios from the original equation into the equation from this statement:

If x = 3, then y = 3 + x = 3 + 3 = 6, so xy = (3)(6) = 18. If y = 0, then x = y – 3 = 0 – 3 = -3, so xy = (-3)(0) = 0. Since there are two possible answers, this statement is not sufficient.

(2) SUFFICIENT: If x3 < 0, then x < 0. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

Ans B

Please someone confirm on the solution, if the approach is correct

Yes, that's correct.

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

\(6xy=x^2y + 9y\) --> \(y(x^2-6x+9)=0\) --> \(y(x-3)^2=0\) --> either \(x=3\) or \(y=0\) (or both).

(1) y – x = 3. If \(y=0\) and \(x=-3\), then \(xy=0\) but if \(x=3\) and \(y=6\), then \(xy=18\). Not sufficient.

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

Re: If 6xy = x^2*y + 9y, what is the value of xy? [#permalink]
19 Jun 2014, 02:10

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Re: If 6xy = x^2*y + 9y, what is the value of xy? [#permalink]
06 Jul 2014, 06:34

Bunuel wrote:

Manhnip wrote:

The equation in question can be rephrased as follows: x^2 y – 6xy + 9y = 0

y(x^2 –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 “What is y?”

(1) INSUFFICIENT: This equation cannot be manipulated or combined with the original equation to solve directly for x or y. Instead, plug the two possible scenarios from the original equation into the equation from this statement:

If x = 3, then y = 3 + x = 3 + 3 = 6, so xy = (3)(6) = 18. If y = 0, then x = y – 3 = 0 – 3 = -3, so xy = (-3)(0) = 0. Since there are two possible answers, this statement is not sufficient.

(2) SUFFICIENT: If x3 < 0, then x < 0. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

Ans B

Please someone confirm on the solution, if the approach is correct

Yes, that's correct.

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

\(6xy=x^2y + 9y\) --> \(y(x^2-6x+9)=0\) --> \(y(x-3)^2=0\) --> either \(x=3\) or \(y=0\) (or both).

(1) y – x = 3. If \(y=0\) and \(x=-3\), then \(xy=0\) but if \(x=3\) and \(y=6\), then \(xy=18\). Not sufficient.

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

Aaah this formatting.. I read the stem as x^(2y) and was totally stuck. Can this ambiguity in notation come in actual test ? I would have expected a bracket or at the least correct ordering like 6xy = yx^2

Re: If 6xy = x^2*y + 9y, what is the value of xy? [#permalink]
06 Jul 2014, 10:12

Expert's post

himanshujovi wrote:

Bunuel wrote:

Manhnip wrote:

The equation in question can be rephrased as follows: x^2 y – 6xy + 9y = 0

y(x^2 –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 “What is y?”

(1) INSUFFICIENT: This equation cannot be manipulated or combined with the original equation to solve directly for x or y. Instead, plug the two possible scenarios from the original equation into the equation from this statement:

If x = 3, then y = 3 + x = 3 + 3 = 6, so xy = (3)(6) = 18. If y = 0, then x = y – 3 = 0 – 3 = -3, so xy = (-3)(0) = 0. Since there are two possible answers, this statement is not sufficient.

(2) SUFFICIENT: If x3 < 0, then x < 0. Therefore, x cannot equal 3, and it follows that y = 0. Therefore, xy = 0.

Ans B

Please someone confirm on the solution, if the approach is correct

Yes, that's correct.

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

\(6xy=x^2y + 9y\) --> \(y(x^2-6x+9)=0\) --> \(y(x-3)^2=0\) --> either \(x=3\) or \(y=0\) (or both).

(1) y – x = 3. If \(y=0\) and \(x=-3\), then \(xy=0\) but if \(x=3\) and \(y=6\), then \(xy=18\). Not sufficient.

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

Aaah this formatting.. I read the stem as x^(2y) and was totally stuck. Can this ambiguity in notation come in actual test ? I would have expected a bracket or at the least correct ordering like 6xy = yx^2

Edited the original post to avoid such confusions in future.

By the way x^2y means x^2*y. If it were \(x^{2y}\) it would be written x^(2y). But don't worry on the actual test the formatting will be clearer.

Re: If 6xy = x^2*y + 9y, what is the value of xy? [#permalink]
06 Jul 2014, 20:23

Bunnuel,

Why is it wrong to divide the original equation by y? Since dividing the equation by y gives you x^2-6x+9=0 which gives you x=3, then statement 1 is sufficient to give you the value of y and therefore give you the value of xy as 18.

Statement 2 is not consistent with the original equation as it shows x^3<0 which is not possible (x=3, so 3^3=27) and is therefore insufficient. Therefore, answer should be A.

Or is it that only because statement 2 is inconsistent with the original equation and therefore we need to try a different approach?

Re: If 6xy = x^2*y + 9y, what is the value of xy? [#permalink]
07 Jul 2014, 00:06

1

This post received KUDOS

Expert's post

ravih wrote:

Bunnuel,

Why is it wrong to divide the original equation by y? Since dividing the equation by y gives you x^2-6x+9=0 which gives you x=3, then statement 1 is sufficient to give you the value of y and therefore give you the value of xy as 18.

Statement 2 is not consistent with the original equation as it shows x^3<0 which is not possible (x=3, so 3^3=27) and is therefore insufficient. Therefore, answer should be A.

Or is it that only because statement 2 is inconsistent with the original equation and therefore we need to try a different approach?

If you divide (reduce) 6xy = x^2*y + 9y by y, you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution (notice that both y = 0 AND x = 3 satisfy the equation).

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero. _________________

Re: If 6xy = x^2*y + 9y, what is the value of xy? [#permalink]
06 May 2015, 00:22

Dear Bunuel, Thanks for the bringing out such fundamental issues.

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

I made the same error: jumped to eliminate y from LHS & RHS > came up with x=3, then solved for y as given in condition 1 to arrive at xy=18 hence right answer = A

Per your explanation: If you divide (reduce) 6xy = x^2*y + 9y by y, you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution (notice that both y = 0 AND x = 3 satisfy the equation).

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

It makes absolute sense and I feel like I am on a totally different wavelength and am solving the problems inconsistently very consistently! This is in-spite of the detailed guides and theoretical resources provided here. Is this just a matter of practice and training one's brain to approach a problem in a more ordered fashion or something fundamentally amiss?

Re: If 6xy = x^2*y + 9y, what is the value of xy? [#permalink]
06 May 2015, 01:15

Expert's post

rohitd80 wrote:

Dear Bunuel, Thanks for the bringing out such fundamental issues.

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

I made the same error: jumped to eliminate y from LHS & RHS > came up with x=3, then solved for y as given in condition 1 to arrive at xy=18 hence right answer = A

Per your explanation: If you divide (reduce) 6xy = x^2*y + 9y by y, you assume, with no ground for it, that y does not equal to zero thus exclude a possible solution (notice that both y = 0 AND x = 3 satisfy the equation).

Never reduce equation by variable (or expression with variable), if you are not certain that variable (or expression with variable) doesn't equal to zero. We can not divide by zero.

It makes absolute sense and I feel like I am on a totally different wavelength and am solving the problems inconsistently very consistently! This is in-spite of the detailed guides and theoretical resources provided here. Is this just a matter of practice and training one's brain to approach a problem in a more ordered fashion or something fundamentally amiss?

Yes, I think practicing similar algebra/inequality questions should help. _________________

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