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Math Help - Need some help

  1. #1
    xfactor0707
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    Exclamation Need some help

    Hi yall im new to this forum but im looking for some help.

    Find all values of the variable for which the expression is undefined.

    1) (4y)/( x- 3)
    2) (4x + 12)/ (x^3 - 9x)
    3) (7)/ (7 -x )
    4) (2y)/ (y^2 - 6y - 7)
    5) (7)/ (x^2 -4)
    6)(4x)/ (2x - 1)

    Can u please help or teach me how 2 do it
    Last edited by xfactor0707; November 18th 2007 at 07:06 AM. Reason: ty jhevon 4 da heads up
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by xfactor0707 View Post
    Hi yall im new to this forum but im looking for some help.

    Find all values of the variable for which the expression is undefined.

    1) 4y/ x- 3 2) 4x + 12/ x^3 - 9x 3) 7/ 7 -x 4) 2y/ y^2 - 6y - 7

    5) 7/ x^2 -4 6) 4x/ 2x - 1

    Can u please help or teach me how 2 do it
    what you wrote is confusing. clarify

    start by writing each new question is a new line

    then be sure to type fractions as follows: (numerator)/(denominator)

    that is, to type \frac {x + 7}{x^2 + 1} type (x + 7)/(x^2 + 1)
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by xfactor0707 View Post
    Hi yall im new to this forum but im looking for some help.

    Find all values of the variable for which the expression is undefined.

    1) (4y)/( x- 3)
    2) (4x + 12)/ (x^3 - 9x)
    3) (7)/ (7 -x )
    4) (2y)/ (y^2 - 6y - 7)
    5) (7)/ (x^2 -4)
    6)(4x)/ (2x - 1)

    Can u please help or teach me how 2 do it
    Much better!

    We are looking for values of x where the expression is undefined. In the case of fractions we look for where the denominator is 0.

    So for 1)
    \frac{4y}{x- 3}
    we know that x - 3 = 0 is forbidden. Thus x = 3 is where the expression is undefined.

    A more complicated example is number 4)
    \frac{2y}{y^2 - 6y - 7}

    We are looking for places where
    y^2 - 6y - 7 = 0

    (y - 7)(y + 1)= 0

    So y cannot be -1 or 7.

    See what you can do with the rest.

    -Dan
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