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Math Help - Simplifying Expression 3 Success

  1. #1
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    Jul 2009
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    Tallinn, Estonia
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    Simplifying Expression 3 Success

    I solved this by my own, using things learned from my previous two threads. I am so proud. I post it here so maybe others can learn from it if they decide to.

    Expression to simplify:
    (\frac{5}{2a+3}+\frac{2}{3-2a}+\frac{2a+9}{4a^2-9})\div\frac{8}{4a^2+12a+9}

    Answer:
    \frac{2a+3}{2}

    Step by step:
    \frac{5}{2a+3}+\frac{2}{3-2a}+\frac{2a+9}{4a^2-9}=\frac{5}{2a+3}-\frac{2}{2a-3}+\frac{2a+9}{4a^2-9}

    {4a^2-9} is formula: a^2-b^2=(a+b)(a-b)
    {4a^2-9}=(2a+3)(2a-3)

    \frac{5}{2a+3}-\frac{2}{2a-3}+\frac{2a+9}{(2a+3)(2a-3)}

    \frac{5(2a-3)-2(2a+3)+(2a+9)}{(2a+3)(2a-3)}

    \frac{10a-15-4a-6+2a+9}{(2a+3)(2a-3)}=\frac{8a-12}{(2a+3)(2a-3)}

    Time to divide,
    4a^2+12a+9 is formula: a^2+2ab+b^2=(a+b)^2
    4a^2+12a+9=(2a+3)^2

    \frac{8a-12}{(2a+3)(2a-3)}\div\frac{8}{(2a+3)^2}

    \frac{8a-12}{(2a+3)(2a-3)}*\frac{(2a+3)^2}{8}

    \frac{8a-12}{(2a-3)}*\frac{(2a+3)}{8}

    \frac{4(2a-3)}{(2a-3)}*\frac{(2a+3)}{8}

    \frac{4}{1}*\frac{(2a+3)}{8}

    \frac{2a+3}{2}

    So good to give something back.
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  2. #2
    Member
    Joined
    Jul 2009
    Posts
    152
    Excellent job. Just one tip:

    noticing that

    \frac{8a-12}{(2a+3)(2a-3)}=\frac{4(2a-3)}{(2a+3)(2a-3)}=\frac{4}{2a+3}

    would have saved you some trouble.

    Cheers.
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