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  1. #1
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    Problem help

    Add or subtract, write each answer in lowest terms.

    2p
    ---
    p - 3

    +

    2+p
    ____
    p

    -
    -6
    ____
    p^2 - 3p


    Where do i begin?
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  2. #2
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    Quote Originally Posted by Tenskypoo View Post
    Add or subtract, write each answer in lowest terms.

    2p
    ---
    p - 3

    +

    2+p
    ____
    p

    -
    -6
    ____
    p^2 - 3p


    Where do i begin?
    Is this what you've tried to write?

    \frac{2p}{p - 3} + \frac{2 + p}{p} - \frac{-6}{p^2 - 3p}?


    If so, you need to multiply each term by a cleverly disguised 1, so that they all have the same denominators. Then you can add the numerators.
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  3. #3
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    Quote Originally Posted by Tenskypoo View Post
    Add or subtract, write each answer in lowest terms.

    2p/(p - 3) + (2 + p)/p - (-6)/(p^2 - 3p)

    Where do i begin?
    To start, factor the third denominator. Then note that the least common denominator will be p(p - 3).

    Convert the three fractions to this common denominator, and then add and subtract the numerators.

    Then simplify the result, if possible, by factoring the numerator and seeing if anything cancels.

    If you get stuck, please reply showing your work, either using LaTeX or else using standard web-safe formatting to make your meaning clear. Your work will probably start something like:

    . . . . . \left(\frac{2p}{p\, -\, 3}\right)\left(\frac{p}{p}\right)\, +\, \left(\frac{2\, +\, p}{p}\right)\left(\frac{p\, -\, 3}{p\, -\, 3}\right)\, +\, \frac{6}{p(p\, -\, 3)}

    Thank you!
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  4. #4
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    Quote Originally Posted by Tenskypoo View Post
    Add or subtract, write each answer in lowest terms.

    2p
    ---
    p - 3

    +

    2+p
    ____
    p

    -
    -6
    ____
    p^2 - 3p


    Where do i begin?
    Quote Originally Posted by Prove It View Post
    Is this what you've tried to write?

    \frac{2p}{p - 3} + \frac{2 + p}{p} - \frac{-6}{p^2 - 3p}?


    If so, you need to multiply each term by a cleverly disguised 1, so that they all have the same denominators. Then you can add the numerators.
    Hi Tenskypoo,

    And the next step...

    \frac{2p}{p - 3} + \frac{2 + p}{p} + \frac{6}{p(p - 3)}

    This makes your LCD = p(p-3)

    Looking at it a little differently than the others, try multiplying this LCD by each term in your expression, cancelling appropriately as you go. By doing this, you create your new numerator.

    p(\rlap{////}p-3)\left(\frac{2p}{\rlap{////}p-3}\right)+\rlap{/}p(p-3)\left(\frac{2+p}{\rlap{/}p}\right)+\rlap{//////}p(p-3)\left(\frac{6}{\rlap{///////}p(p-3)}\right) \Longrightarrow 2p^2+(p-3)(p+2)+6

    Expand and combine terms and place this over p(p-3)

    \frac{2p^2+(p-3)(p+2)+6}{p(p-3)}

    Now, simplify.

    Just a variation on a theme.
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  5. #5
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    Quote Originally Posted by masters View Post
    Hi Tenskypoo,

    And the next step...

    \frac{2p}{p - 3} + \frac{2 + p}{p} + \frac{6}{p(p - 3)}

    This makes your LCD = p(p-3)

    Looking at it a little differently than the others, try multiplying this LCD by each term in your expression, cancelling appropriately as you go. By doing this, you create your new numerator.

    p(\rlap{////}p-3)\left(\frac{2p}{\rlap{////}p-3}\right)+\rlap{/}p(p-3)\left(\frac{2+p}{\rlap{/}p}\right)+\rlap{//////}p(p-3)\left(\frac{6}{\rlap{///////}p(p-3)}\right) \Longrightarrow 2p^2+(p-3)(p+2)+6

    Expand and combine terms and place this over p(p-3)

    \frac{2p^2+(p-3)(p+2)+6}{p(p-3)}

    Now, simplify.

    Just a variation on a theme.
    You can't just multiply terms by anything unless you have an equation you can do it to both sides of.

    The only thing you can multiply each term by is 1.
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  6. #6
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    Quote Originally Posted by Prove It View Post
    You can't just multiply terms by anything unless you have an equation you can do it to both sides of.

    The only thing you can multiply each term by is 1.
    Hi Prove It,

    I understand what you are saying. Really, I do.

    The only contribution I wanted to make here is that if you want to find a new numerator given a different denominator (LCD), then you can do what I just demonstrated. It's just a technique.

    For instance, say, you need to convert \frac{3}{4} to \frac{?}{8}, you would just multiply \frac{3}{4} by 8 to get 6. And that's your new numerator.
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