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Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]
15 Oct 2012, 03:22

3

This post received KUDOS

Expert's post

Jp27 wrote:

If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?

A. (18-3y)/y^3 B. 18/y^2 C. 18/(y^2+3y) D. 9/y^2 E. 36/y^2

\((xy)^2+3(xy)-18=0\) --> solving for \(xy\) --> \(xy=-6\) (not a valid solution: \(xy\) must be positive as both unknowns are positive) or \(xy=3\) --> so \(xy=3\) --> \(x=\frac{3}{y}\) --> \(x^2=\frac{9}{y^2}\).

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]
15 Oct 2012, 03:38

Bunuel wrote:

Jp27 wrote:

If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?

A. (18-3y)/y^3 B. 18/y^2 C. 18/(y^2+3y) D. 9/y^2 E. 36/y^2

\((xy)^2+3(xy)-18=0\) --> solving for \(xy\) --> \(xy=-6\) (not a valid solution: \(xy\) must be positive as both unknowns are positive) or \(xy=3\) --> so \(xy=3\) --> \(x=\frac{3}{y}\) --> \(x^2=\frac{9}{y^2}\).

Answer: D.

Hope it's clear.

Hi Bunuel - thanks for your swift response.

Where I can get similar question like these for practice?

I have done the roots and equation set in your signature, are there any other sets that I can use for practice? Sincere thanks!

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]
15 Oct 2012, 03:44

1

This post received KUDOS

Expert's post

Jp27 wrote:

Bunuel wrote:

Jp27 wrote:

If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?

A. (18-3y)/y^3 B. 18/y^2 C. 18/(y^2+3y) D. 9/y^2 E. 36/y^2

\((xy)^2+3(xy)-18=0\) --> solving for \(xy\) --> \(xy=-6\) (not a valid solution: \(xy\) must be positive as both unknowns are positive) or \(xy=3\) --> so \(xy=3\) --> \(x=\frac{3}{y}\) --> \(x^2=\frac{9}{y^2}\).

Answer: D.

Hope it's clear.

Hi Bunuel - thanks for your swift response.

Where I can get similar question like these for practice?

I have done the roots and equation set in your signature, are there any other sets that I can use for practice? Sincere thanks!

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]
16 Dec 2013, 18:21

Bunuel wrote:

Jp27 wrote:

If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?

A. (18-3y)/y^3 B. 18/y^2 C. 18/(y^2+3y) D. 9/y^2 E. 36/y^2

\((xy)^2+3(xy)-18=0\) --> solving for \(xy\) --> \(xy=-6\) (not a valid solution: \(xy\) must be positive as both unknowns are positive) or \(xy=3\) --> so \(xy=3\) --> \(x=\frac{3}{y}\) --> \(x^2=\frac{9}{y^2}\).

Answer: D.

Hope it's clear.

very elegant&simple...if you spot one variable in two like it _________________

Good things come to those who wait… greater things come to those who get off their ass and do anything to make it happen...

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]
11 Jun 2015, 02:36

Expert's post

noTh1ng wrote:

I don't understand why you can not divide by y^2 here? Pls help

If you divide x^2 * y^2 = 18 – 3xy by y^2 you get x^2 = 18/y^2 - 3/y = (18-3y)/y^2. We don't have (18-3y)/y^2 among answer choices, thus we need other approach. _________________

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]
11 Jun 2015, 03:34

Bunuel wrote:

noTh1ng wrote:

I don't understand why you can not divide by y^2 here? Pls help

If you divide x^2 * y^2 = 18 – 3xy by y^2 you get x^2 = 18/y^2 - 3/y = (18-3y)/y^2. We don't have (18-3y)/y^2 among answer choices, thus we need other approach.

I sometimes don't get the GMAC..

just because it's not among the answer choices doesnt make it wrong imho

Re: If x and y are positive and x^2y^2 = 18 – 3xy, then x^2 ? [#permalink]
11 Jun 2015, 20:45

1

This post received KUDOS

Expert's post

noTh1ng wrote:

Bunuel wrote:

noTh1ng wrote:

I don't understand why you can not divide by y^2 here? Pls help

If you divide x^2 * y^2 = 18 – 3xy by y^2 you get x^2 = 18/y^2 - 3/y = (18-3y)/y^2. We don't have (18-3y)/y^2 among answer choices, thus we need other approach.

I sometimes don't get the GMAC..

just because it's not among the answer choices doesnt make it wrong imho

If an expression is not in the options, it doesn't make the expression wrong. You might just need to manipulate it further.

Here, dividing by y^2 doesn't work: \(x^2 * y^2 = 18 – 3xy\) When you divide by y^2, you get \(x^2 = 18/y^2 – 3x/y\) How do you separate the x and y since you need to write x^2 in terms of y only? You will need to divide by xy and then solve for it.

By the way, GMAC is absolutely reasoning based. If something confuses you, especially in Quant, it means you have missed a point. _________________

My last interview took place at the Johnson School of Management at Cornell University. Since it was my final interview, I had my answers to the general interview questions...