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skinsvt
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0<X<Y What does (X+Y)^2/(X-Y)^2 = (1) X^2 +Y^2 = 3XY 13 May 2007, 16:50
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0<X<Y

What does (X+Y)^2/(X-Y)^2 =

(1) X^2 +Y^2 = 3XY

(2) XY = 3


I'm getting B for the answer, but the official answer according to GMAT Prep is A. Could some one explain?

I was thinking x^2+2xy+y^2/x^2-2xy+y^2 = 4xy = 4(3) =12 by using statement 2.

Not sure how the answer is A.



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0<X<Y What does (X+Y)^2/(X-Y)^2 = (1) X^2 +Y^2 = 3XY Updated on: 13 May 2007, 19:20
Originally posted by Sergey_is_cool on 13 May 2007, 17:16.
Last edited by Sergey_is_cool on 13 May 2007, 19:20, edited 1 time in total.
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skinsvt
0<X<Y

What does (X+Y)^2/(X-Y)^2 =

(1) X^2 +Y^2 = 3XY

(2) XY = 3


I'm getting B for the answer, but the official answer according to GMAT Prep is A. Could some one explain?

I was thinking x^2+2xy+y^2/x^2-2xy+y^2 = 4xy = 4(3) =12 by using statement 2.

Not sure how the answer is A.
Show more


You were moving in the right direction

(1) X^2 +Y^2 = 3XY => we can rewrite the equation x^2+2xy+y^2/x^2-2xy+y^2 as (x^2+y^2)+2xy/(x^2+y^2)-2xy= 3xy+2xy/3xy-2xy= 5xy/xy=5

statement (2) is not sufficient because we need at least one more equation to solve it.
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0<X<Y What does (X+Y)^2/(X-Y)^2 = (1) X^2 +Y^2 = 3XY 13 May 2007, 19:17
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I seem to be messing up with different types of problems. I think I got it now, but to make sure had the problem been:

2^(x+y)^2/2^(x-y)^2 then B would have been the correct answer.

2^(x^2 +2xy + y^2 - x^2 +2xy-y^2) = 2^4xy = 2^4(3) = 2^12


But since the problem was (x+y)^2/(x-y)^2 then

we have x^2 +2xy +y^2/x^2 - 2xy +y^2 and cannot simplify to 4xy but it would simplify to 4Y^2. I think this is where I'm confused because when we have this same formula as part of an exponent as we do above I want to cancel everything out and have 4xy as the remainder.


and had the problem been 2^(X+Y)/2^(X-Y)= 2^6 solve for Y
we would cancel out the 2's and have X+Y/X-Y = 6
and then have 2Y=6 and Y=3.

Sorry for the long post. Just want to make sure I'm understanding everything right. Could someone please clarify for me? Thanks
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0<X<Y What does (X+Y)^2/(X-Y)^2 = (1) X^2 +Y^2 = 3XY 14 May 2007, 07:25
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I seem to be messing up with different types of problems. I think I got it now, but to make sure had the problem been:

2^(x+y)^2/2^(x-y)^2 then B would have been the correct answer.

2^(x^2 +2xy + y^2 - x^2 +2xy-y^2) = 2^4xy = 2^4(3) = 2^12


But since the problem was (x+y)^2/(x-y)^2 then

we have x^2 +2xy +y^2/x^2 - 2xy +y^2 and cannot simplify to 4xy but it would simplify to 4Y^2. I think this is where I'm confused because when we have this same formula as part of an exponent as we do above I want to cancel everything out and have 4xy as the remainder.


and had the problem been 2^(X+Y)/2^(X-Y)= 2^6 solve for Y
we would cancel out the 2's and have X+Y/X-Y = 6
and then have 2Y=6 and Y=3.

Sorry for the long post. Just want to make sure I'm understanding everything right. Could someone please clarify for me? Thanks
Show more


x^2 +2xy +y^2/x^2 - 2xy +y^2

using statement (1) we replace X^2 +Y^2 with 3XY and after that solve the equation.

Answer is A
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0<X<Y What does (X+Y)^2/(X-Y)^2 = (1) X^2 +Y^2 = 3XY 14 May 2007, 07:32
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Sorry Sergey..Thanks for your patience.


What if the problem was..


x + 2xy + y/ x - 2xy + y

Are we able to cancel out the x's and y's to get 4xy or must we factor first in order to cancel?
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0<X<Y What does (X+Y)^2/(X-Y)^2 = (1) X^2 +Y^2 = 3XY 14 May 2007, 07:57
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Sorry Sergey..Thanks for your patience.


What if the problem was..


x + 2xy + y/ x - 2xy + y

Are we able to cancel out the x's and y's to get 4xy or must we factor first in order to cancel?
Show more


If we have an equation x + 2xy + y/ x - 2xy + y and
statement (1) says x+y=3xy for example than we would be able to substitute
(x+y) with 3xy and rewrite the equation as 3xy+2xy/3yx-2xy and get
5xy/xy where we can cancel out xy and get 5 as an answer

Let me know if it makes sense.
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0<X<Y What does (X+Y)^2/(X-Y)^2 = (1) X^2 +Y^2 = 3XY 14 May 2007, 08:35
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Yeah- I think I understand this problem, but I am now a little confused with other problems that look similar. This problem just triggered my confusion. I think you've helped to answer my questions by talking through it, but I want to make sure if the equations were different. For example,

x-y/x+y = 6

In the equation above we cancel out the x's and get 2y=6.

When we have a problem like the one we've been discussing in previous posts, my gut tells me to cancel out the x's and cancel out the y's (similar to the problem above except we have x^2 and 2xy)) and then take 2xy - (-2xy) to get 4xy. Guess I'm asking...can we only factor out x's and y's when we have single variable as above?

So if we had and wanted to solve for Y

x^2 + y^2/ x^2 -y^2 = 24

I know this can be factored out and we solve for y.
(x+y)(x+y)/(x+y)(x-y) = 24
= 2y=24 y=12

but when we add 2xy to the equation directly above my gut is telling me to subtract similar to the approach taken when we have the same equation as a power

2^(x+y)^2/2^(x-y)^2

Here we distribute and subtract to get 2^4xy. Sorry if I'm not making any sense..just trying to decipher between different formulas, factoring, canceling etc. I think I understand, just want clarification from someone who is more familiar.
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0<X<Y What does (X+Y)^2/(X-Y)^2 = (1) X^2 +Y^2 = 3XY 14 May 2007, 09:10
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Yeah- I think I understand this problem, but I am now a little confused with other problems that look similar. This problem just triggered my confusion. I think you've helped to answer my questions by talking through it, but I want to make sure if the equations were different. For example,

x-y/x+y = 6

In the equation above we cancel out the x's and get 2y=6.

When we have a problem like the one we've been discussing in previous posts, my gut tells me to cancel out the x's and cancel out the y's (similar to the problem above except we have x^2 and 2xy)) and then take 2xy - (-2xy) to get 4xy. Guess I'm asking...can we only factor out x's and y's when we have single variable as above?

So if we had and wanted to solve for Y

x^2 + y^2/ x^2 -y^2 = 24

I know this can be factored out and we solve for y.
(x+y)(x+y)/(x+y)(x-y) = 24
= 2y=24 y=12

but when we add 2xy to the equation directly above my gut is telling me to subtract similar to the approach taken when we have the same equation as a power

2^(x+y)^2/2^(x-y)^2

Here we distribute and subtract to get 2^4xy. Sorry if I'm not making any sense..just trying to decipher between different formulas, factoring, canceling etc. I think I understand, just want clarification from someone who is more familiar.
Show more


x-y/x+y = 6 you can't cancel x's out because it's not a multiplication
you can cancel out only when there is * or /
for example 2xy/3x= 2y/3

So if we had and wanted to solve for Y

x^2 + y^2/ x^2 -y^2 = 24

(x+y)(x+y)/(x+y)(x-y) = 24 we can cancel only (x+y) and get x+y/x-y=24 and we'd need one more equation to solve it.
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0<X<Y What does (X+Y)^2/(X-Y)^2 = (1) X^2 +Y^2 = 3XY 14 May 2007, 11:01
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Thanks for your patience...I think I've got it now. I've been working with Exponent formula only- adding subtracting, etc for the last couple of weeks and had forgotten some basic non exponent formula rules.
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0<X<Y What does (X+Y)^2/(X-Y)^2 = (1) X^2 +Y^2 = 3XY 14 May 2007, 12:19
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Sorry Sergey..Thanks for your patience.


What if the problem was..


x + 2xy + y/ x - 2xy + y

Are we able to cancel out the x's and y's to get 4xy or must we factor first in order to cancel?
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you can't factor x + 2xy + y/ x - 2xy + y !!

if you try and factor with x you get:

x(1+2y) + y/x(1-2y) + y

and if you try and factor with y you get:

x + (2x+1)/x-(2x+1) + x

:)



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