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Math Help - Solve X and Y problem

  1. #1
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    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?
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by thecapaccino View Post
    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.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by thecapaccino View Post
    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
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    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
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  5. #5
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    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?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by thecapaccino View Post
    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
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by thecapaccino View Post
    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)]

    in your original problem.

    -Dan
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  8. #8
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    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?
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  9. #9
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by thecapaccino View Post
    [(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
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