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Math Help - How do I simplify these polynomial fractions?

  1. #1
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    How do I simplify these polynomial fractions?

    I think I might have gotten the answer to these but I don't know....

    (9x^2-3x)/3x

    answer
    3x-1
    ^^^^^^^^^^^

    (x^2-5x+4/x-1

    answer
    x-4
    ^^^^^^^^^^^^^^^^

    (x-y)/(x+y)+((x+y)/(x-y)

    answer
    (x-y)+(x+y)

    ^^^^^^^^^^^^^^^^

    One last thing I'm unsure of is a question that asks me to solve an equation. I have the answer, I used logic to find it and now I can't figure out how to show working for it.

    1/x-1/(4-x)=1/3

    x=2
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  2. #2
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    Quote Originally Posted by quikwerk View Post
    I think I might have gotten the answer to these but I don't know....

    (9x^2-3x)/3x

    answer
    3x-1
    ^^^^^^^^^^^ Correct

    (x^2-5x+4/x-1

    answer
    x-4
    ^^^^^^^^^^^^^^^^ Correct

    (x-y)/(x+y)+((x+y)/(x-y)

    answer
    (x-y)+(x+y)

    ^^^^^^^^^^^^^^^^ No. It's  \frac{2(x^2 + y^2)}{x^2 - y^2}

    One last thing I'm unsure of is a question that asks me to solve an equation. I have the answer, I used logic to find it and now I can't figure out how to show working for it.

    1/x-1/(4-x)=1/3

    x=2

    Try this:
     \frac{1}{x} \times \frac{4-x}{4-x} - \frac{1}{4-x} \times \frac{x}{x} = \frac{1}{3}

     \frac{4-2x}{4x - x^2} = \frac{1}{3}

     12 - 6x = 4x - x^2

     x^2 - 10x + 12 = 0

    Solve this using the quadratic equation to get  5 \pm \sqrt{13}
    .
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  3. #3
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    Thanks for your help, but I think I typed out that last question badly it was supposed to read: 1/x-1/x+4 NOT "1/x-1/(4-x)=1/3".....
    My bad, I'm really sorry.
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  4. #4
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    Quote Originally Posted by quikwerk View Post
    Thanks for your help, but I think I typed out that last question badly it was supposed to read: 1/x-1/x+4 NOT "1/x-1/(4-x)=1/3".....
    My bad, I'm really sorry.
    Same idea,

     \left( \frac{1}{x} \times \frac{x+4}{x+4} \right)+ \left( \frac{1}{x+4} \times \frac{x}{x} \right) = \frac{1}{3}

     \frac{x+4 - x}{x(x+4)} = \frac{1}{3}

     12 = x^2 + 4x

     x^2 + 4x - 12 = 0

     (x+6)(x-2) = 0

    So solutions are x = -6 and x = 2
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