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

if (x2y) in (6xy = x2y + 9y) is x^2y, x can not be -ve.

6xy = x^2 + 9y
6xy = y (x^2 + 9)
x^2 - 6x + 9 = 0
(x - 3)^2 = 0
so x = 3.

1: y = 6. sufficient.
2: absurd. nsf.

probably A but the question is not a formulated properly.

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.

Hope it helps.

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.

Thats an important lesson! thanks Bunuel!

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.

Bunuel, I did but there is some minor point that I am just not getting. For instance:
if x3 < 0 i.e. x is negative, how do we know that y will be 0?
I am not following how either the original equation - y(x+3)^2 =0 - or the above solutions are leading us to this conclusion.
thanks a ton

For y(x−3)^2 to be 0, either x must be 3 or y must be 0 (or both). x^3 < 0 implies that x is NOT 3, thus y MUST be 0 (in order y(x−3)^2 to be 0).

Now, I get it. Thank you!

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) y – x = 3

(2) x^3 < 0

If we modify the question, 0=x^2*y-6xy+9y, or 0=y(x^2-6x+9), or 0=y(x-3)^2, and we want to know whether y=0 or x=3. There are 2 variables (x,y) and 1 equation (0=x^2*y-6xy+9y). 2 more equations are given from the 2 conditions, so there is high chance (D) will be our answer.
From condition 1, y=0 and x=-3 or x=3 and y=6. This is not sufficient as this is not unique.
From condition 2, x^3<0--> x<0. But x=3>0, and y has to be y=0, and xy=0, so the condition is sufficient; the answer is (B).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

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
“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.
The correct answer is B.

This video is not only the video solution of this question, but also an explanation how the DS questions should be approached.

It also highlights a common mistake of many test takers

it is given in the statement x=+- 3,y=0.
then in statement 1, why we are taking different value (y=6) . Is it to satisfy the equation?

bcz as per my ans D should be correct.

Bunuel's approach is the best, but here is an alternative solution.
Rearrange the equation in the stem to \(xy = \frac{y(x^2+9)}{6}\) and now you're looking for the value of xy

(1) \(y-x=3\) so \(y=3+x\) > plug this into the equation in the stem
\(x(3+x)=\frac{(3+x)(x^2+9)}{6}\)
\(3x+x^2 = \frac{3x^2+27+x^3+9x}{6}\)
\(18x+6x^2=3x^2+27+x^3+9x\)
\(x^3-3x^2-9x+27=0\)
\(x(x^2-9)-3(x^2-9)=0\)
\((x+3)(x-3)^2=0\)
x=3, -3
If x=3, y=6 and xy=18. If x=-3, y=0 and xy=0. NOT SUFFICIENT.

(2) \(x^3<0\)
Notice that if x is negative then y must be equal to zero because in the equation \(xy = \frac{y(x^2+9)}{6}\) x will be negative on the LHS while (x^2+9) will be positive. Since a negative cannot equal a positive, y must equal zero to make the equation valid. SUFFICIENT

Why can't we eliminate y from LHS and RHS? we usually eliminate variables like that.