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

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).

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.

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.

Answer (D)

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

[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
Dinesh

VeritasPrepKarishma Bunuel niks18 amanvermagmat chetan2u



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)

Hi adkikani

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 \(1+y-1=0 => y=0\)

niks18

Thanks for prompt support.


Correct, we are on same page.



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.

Hi adkikani

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.

Given: \(y(x-3)^2 = 0\)

it is not y^2 but x^2 so this Binomial theorem does not apply

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