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?

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- April 17th 2007, 07:47 PMthecapaccinoSolve 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? - April 17th 2007, 08:22 PMecMathGeek
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. - April 17th 2007, 08:29 PMJhevon
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 - April 17th 2007, 08:30 PMJhevon
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

- April 17th 2007, 08:32 PMthecapaccinoRe-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? - April 17th 2007, 08:35 PMJhevon
- April 18th 2007, 04:41 AMtopsquark
- April 18th 2007, 04:46 AMthecapaccinoMy 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? - April 18th 2007, 04:51 AMtopsquark