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Math Help - A couple of algebra problems...

  1. #1
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    A couple of algebra problems...

    Hi, I'm new to the board.

    I had a big ol' heaping pile of algebra homework to do recently, and I'm happy to say that I've actually figured most of it out, save for these four problems. It's really just two basic ideas that I don't understand, as the problems come in pairs.

    For the first pair, I'm just asked to simplify:

    a^2-2ab+b^2
    ______________

    a-b


    and...


    a^2-2ab+b^2
    ______________

    a+b

    I've been trying to solve these using polynomial long division, but I don't think I'm doing the descending-power reordering right or something.

    The second pair are just addition and subtraction problems, but I'm stumped.

    a-1
    ____

    a+1

    +

    a+1
    ____

    a-1



    and



    4
    ____

    a-5

    -

    1
    ____

    5-a

    If someone can help me understand how to solve these problems, I'd appreciate it.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by Grimes View Post
    Hi, I'm new to the board.

    I had a big ol' heaping pile of algebra homework to do recently, and I'm happy to say that I've actually figured most of it out, save for these four problems. It's really just two basic ideas that I don't understand, as the problems come in pairs.

    For the first pair, I'm just asked to simplify:

    \frac{a^2-2ab+b^2}{a-b}
    First, factor the numerator.

    \frac{(a-b)(a-b)}{a-b}

    Then, divide out any common factors. In this case, a-b is common in both the numerator and denominator. So, we end up with:

    \boxed{a-b}

    Quote Originally Posted by Grimes View Post
    \frac{a^2-2ab+b^2}{a+b}
    Do this one the same way.

    \frac{(a-b)(a-b)}{a+b}

    There are no common factors here, so nothing cancels out. This cannot be simplified further.

    Quote Originally Posted by Grimes View Post

    I've been trying to solve these using polynomial long division, but I don't think I'm doing the descending-power reordering right or something.

    The second pair are just addition and subtraction problems, but I'm stumped.

    \frac{a-1}{a+1}+\frac{a+1}{a-1}
    We need to find the common denominator. Since they're both different the common denominator will be (a+1 )(a-1)

    Now rewrite each fraction using the new denominator. We do this by multiplying this common denominator by each term; then put that result over the common denominator. Like this...

    \frac{(a+1 )(a-1)\left(\frac{a-1}{a+1}\right)+(a+1 )(a-1)\left(\frac{a+1}{a-1}\right)}{(a+1)(a-1)}=\frac{(a-1)(a-1)+(a+1)(a+1)}{(a+1)(a-1)}=

    \frac{a^2-2a+1+a^2+2a+1}{(a+1)(a-1)}=\frac{2a^2+2}{(a+1)(a-1)}=\frac{2(a^2+1)}{(a+1)(a-1)}

    Quote Originally Posted by Grimes View Post

    \frac{4}{a-5}-\frac{1}{5-a}
    Let's rearrange that second fraction to get both denominators to look the same.

    \frac{4}{a-5}-\frac{1}{-(a-5)}=\frac{4}{a-5}+\frac{1}{a-5}=\frac{5}{a-5}
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  3. #3
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    Thanks a lot for the help!
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