# Solve X and Y problem

• Apr 17th 2007, 07:47 PM
thecapaccino
Solve X and Y problem
If x=2 and xy does not =0, what is the value of

x^4*y^4 - (xy)^2 / x^3 * y^2

My answer is 3y^2 / 2y^2...but not sure if that is the answer or should I just say 3/2?
• Apr 17th 2007, 08:22 PM
ecMathGeek
Quote:

Originally Posted by thecapaccino
If x=2 and xy does not =0, what is the value of

x^4*y^4 - (xy)^2 / x^3 * y^2

My answer is 3y^2 / 2y^2...but not sure if that is the answer or should I just say 3/2?

I assume this problem is
(x^4*y^4 - x^2*y^2)/(x^3*y^2)

I'm not sure where you got 3y^2 / 2y^2, but I don't see that you can get that from this.

If we break up the fraction, we get
(x^4*y^4)/(x^3*y^2) - (x^2*y^2)/(x^3*y^2) = x*y^2 - 1/x

Maybe I'm misunderstanding the problem.
• Apr 17th 2007, 08:29 PM
Jhevon
Quote:

Originally Posted by thecapaccino
If x=2 and xy does not =0, what is the value of

x^4*y^4 - (xy)^2 / x^3 * y^2

My answer is 3y^2 / 2y^2...but not sure if that is the answer or should I just say 3/2?

yes, if that is the answer, you should say 3/2, but how did you get that answer? am i misinterpreting the problem? please type clearly, use brackets to show what is being divided by what.

i think the problem is:

[x^4*y^4 - (xy)^2]/(x^3*y^2)
= [x^4*y^4 - x^2*y^2]/(x^3*y^2)
= [(x^2*y^2)(x^2*y^2 - 1)]/(x^3*y^2)
= (x^2*y^2 - 1)/(x)
= (4y^2 - 1)/2

i plugged in x = 2 in the final step. i didn't split the fraction up as ecMathGeek did
• Apr 17th 2007, 08:30 PM
Jhevon
it is a frustrating waste of time when we have to be guessing what the problem is, please, for our sake and yours, state your problems clearly. use parenthesis
• Apr 17th 2007, 08:32 PM
thecapaccino
Re-write
Okay...let me try to re-write again. There was a slight typo on the first y^2.

If x=2 and xy does not =0, what is the value of

(x^4)*(y^2) - (xy)^2 / (x^3) * (y^2) =

I plugged in x to get the following

16*(y^2) - 4*(y^2) / 8*(y^2)
then

12*(y^2) / 8*(y^2)
then reduce

3(y^2) / 2(y^2)

Is that still wrong?
• Apr 17th 2007, 08:35 PM
Jhevon
Quote:

Originally Posted by thecapaccino
Okay...let me try to re-write again. There was a slight typo on the first y^2.

If x=2 and xy does not =0, what is the value of

(x^4)*(y^2) - (xy)^2 / (x^3) * (y^2) =

I plugged in x to get the following

16*(y^2) - 4*(y^2) / 8*(y^2)
then

12*(y^2) / 8*(y^2)
then reduce

3(y^2) / 2(y^2)

Is that still wrong?

well, in that case, you are correct. write it as 3/2
• Apr 18th 2007, 04:41 AM
topsquark
Quote:

Originally Posted by thecapaccino
Okay...let me try to re-write again. There was a slight typo on the first y^2.

If x=2 and xy does not =0, what is the value of

(x^4)*(y^2) - (xy)^2 / (x^3) * (y^2) =

I plugged in x to get the following

16*(y^2) - 4*(y^2) / 8*(y^2)
then

12*(y^2) / 8*(y^2)
then reduce

3(y^2) / 2(y^2)

Is that still wrong?

That is incorrect.
16*(y^2) - 4*(y^2) / 8*(y^2)

= 16y^2 - (4/8)y^2 = 16y^2 - (1/2)y^2 = (31/2)y^2

Unless you meant:

[(x^4)*(y^2) - (xy)^2] / [(x^3) * (y^2)]

-Dan
• Apr 18th 2007, 04:46 AM
thecapaccino
My fault (again) for forgetting extra parenthesis
[(x^4)*(y^2) - (xy)^2] / [(x^3) * (y^2)] - indeed was what I meant.

I am so sorry for the confusion...I should have put the outside set of parentheses on the numerator problem.

So...bearing that in mind...my answer is correct?
• Apr 18th 2007, 04:51 AM
topsquark
Quote:

Originally Posted by thecapaccino
[(x^4)*(y^2) - (xy)^2] / [(x^3) * (y^2)] - indeed was what I meant.

I am so sorry for the confusion...I should have put the outside set of parentheses on the numerator problem.

So...bearing that in mind...my answer is correct?

Given that as the form, then I agree with Jhevon, you are correct. :) (Since xy is not zero then y is not zero so you can legally cancel the y^2.)

-Dan