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Math Help - Can you please solve these for me?

  1. #1
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    Can you please solve these for me?

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  2. #2
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    Quote Originally Posted by TheNotoriousVAG View Post
    Note that if two fractions have the same denominator they may be combined.

    Mathematically: \frac{a}{c} \pm \frac{b}{c} = \frac{a\pm b}{c}\: ,\: c \neq 0

    Difference of two squares: a^2-b^2 = (a-b)(a+b)

    1. \frac{5}{y-2} = y+2

    Multiply each side by y-2 to clear the denominator (note that the right side will be the difference of two squares so it'd be easier to expand). Then take 5 from each side and solve the quadratic (this will also be the difference of two squares)


    2. \frac{x}{3} - \frac{2}{3} = \frac{1}{x}

    You can then combine the left side and multiply through by 3x to give

    x^2-2x = 3 and solve the quadratic using your favourite method

    3. \frac{y+2}{y} = \frac{1}{y-5}

    Cross multiply to give (y+2)(y+5) = y. Expand and solve.

    4. \frac{x^2}{x-4} - \frac{7}{x-4} = 0

    Combine denominator to give: \frac{x^2-7}{x-4} = 0

    In this case x-4 will cancel (but remember that x=4 cannot be a solution). x^2-7=0 can be solved using the difference of two squares.

    5. \frac{x^2}{x+3} - \frac{5}{x+3} = 0

    Do the same as in question 4 but solve x^2-5=0 instead.
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  3. #3
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    thanks man

    Can you give me the answer to #3 I can't seem to get it right.

    Can you give me the answer to #4 and 5 too. I don't get what u mean
    Last edited by mr fantastic; May 10th 2009 at 08:00 PM. Reason: Merged posts
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  4. #4
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    Quote Originally Posted by TheNotoriousVAG View Post
    Can you give me the answer to #3 I can't seem to get it right.
    \frac {y+2}{y} = \frac {1}{y-5}

    y+2 = y(\frac {1}{y-5})

    y+2 = \frac {y}{y-5}

    (y-5)(y+2) = y

    y^2 +2y - 5y -10 = y

    y^2 - 3y -10 = y

    y^2 - 4y -10 = 0

    Find y using the quadratic formula : y = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}
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  5. #5
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    Quote Originally Posted by TheNotoriousVAG View Post
    Can you give me the answer to #4 and 5 too. I don't get what u mean
    e^(i*pi) explained exactly what you had to do, unfortunately, we are not here to provide answers but tips and paths for you to find your own.
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  6. #6
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    Do i have to use the quadratic formula for 4 and 5?
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  7. #7
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    Quote Originally Posted by TheNotoriousVAG View Post
    Do i have to use the quadratic formula for 4 and 5?
    If you can transform the equation into ax^2 + bx + c = 0 form, then, yes.
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