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Math Help - Working with expressions.

  1. #1
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    Working with expressions.

    Please, if anyone could solve and explain to me the process of solving these two problems I would be extremely grateful. (It's not about the result, I really need to know HOW to do this, so If I can see a process of solving these problems I hope I will be able to understand it).

    1) IF (b-2a)/(4a+3b)=2 (a,b!=0, 4a+3b!=0), THEN (2a^2-3ab+5b^2)/(4a^2+3b^2) EQUALS?

    2) Value of the expression (25^(0,3)*5^(0,4))/12^(-1/3) belongs to the interval (which)?
    Last edited by mr fantastic; March 11th 2011 at 01:16 PM. Reason: Re-titled.
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  2. #2
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    Hi
    Quote Originally Posted by ikislav View Post
    1) IF (b-2a)/(4a+3b)=2 (a,b!=0, 4a+3b!=0), THEN (2a^2-3ab+5b^2)/(4a^2+3b^2) EQUALS?
    Since b \neq 0 you can divide by b

    Let q = \frac{a}{b}

    \frac{b-2a}{4a+3b}=2=\frac{1-2q}{3+4q}
    from which you can get the value for q that you can then plug into

    \frac{2a^2-3ab+5b^2}{4a^2+3b^2}=\frac{2q^2-3q+5}{4q^2+3}
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  3. #3
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    Since I have started writing,

    For 1), the advice above is equivalent to expressing b through a from the first equation and substituting it into the second expression.

    For 2), does 0,3 mean 3/10 or an interval from 1 to 3? In the English-speaking world, period is most commonly used as a decimal mark. If it means 3/10, then the value of this expression belongs to many intervals. For example, 1\le 25^{0.3}\cdot5^{0.4}/12^{-1/3}=25^{0.3}\cdot5^{0.4}\cdot12^{1/3}<25\cdot5\cdot12. To get a tighter upper bound, 25^{0.3}\cdot5^{0.4}\cdot12^{1/3}<25^{1/3}\cdot5^{1/2}\cdot12^{1/3}=\sqrt[3]{25\cdot12}\sqrt{5}<7\cdot2.5=15.
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  4. #4
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    Quote Originally Posted by running-gag View Post
    Hi


    Since b \neq 0 you can divide by b

    Let q = \frac{a}{b}

    \frac{b-2a}{4a+3b}=2=\frac{1-2q}{3+4q}
    from which you can get the value for q that you can then plug into

    \frac{2a^2-3ab+5b^2}{4a^2+3b^2}=\frac{2q^2-3q+5}{4q^2+3}
    Ok, I understand the principle... thank you very much.
    But I dont understand the following: If I divide the second expression by b, how do I get from -3ab to -3q? Wouldn't it be -3a?
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  5. #5
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    You should divide the second expression by b^2.
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  6. #6
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    Quote Originally Posted by emakarov View Post
    You should divide the second expression by b^2.
    Yes, I understand now. Thank you.
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