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Math Help - These Exponents Don't Work Out

  1. #1
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    These Exponents Don't Work Out

    5a/6bc^2 + 3b/8a^2c. The LCM for this, as I worked it out, is 24a^2bc^3. The book say's it's 24a^2bc^2. And my simplification was: 20a^3 + 9b^2c^3/24a^2bc^3. The book's answer is 20a^3 + 9b^2c/24a^2bc^2. Why isn't the LCM c^3? And why is there only one "c" variable in the simplification (i.e. 20a^3 + 9b^2c). Why are the "c" variables being cancelled out while the "b" and "a" variables aren't?

    Thanks guys.
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    Re: These Exponents Don't Work Out

    Quote Originally Posted by MathClown View Post
    5a/6bc^2 + 3b/8a^2c. The LCM for this, as I worked it out, is 24a^2bc^3. The book say's it's 24a^2bc^2. And my simplification was: 20a^3 + 9b^2c^3/24a^2bc^3. The book's answer is 20a^3 + 9b^2c/24a^2bc^2. Why isn't the LCM c^3? And why is there only one "c" variable in the simplification (i.e. 20a^3 + 9b^2c). Why are the "c" variables being cancelled out while the "b" and "a" variables aren't?

    Thanks guys.
    First of all, this is VERY hard to read, some brackets would help. I assume that your expression is \displaystyle \begin{align*} \frac{5a}{6bc^2} + \frac{3b}{8a^2c} \end{align*}, if you look at the factors in each denominator, you have \displaystyle \begin{align*} 2 \cdot 3 \cdot b \cdot c \cdot c \end{align*}, and in the second you have \displaystyle \begin{align*} 2 \cdot 2 \cdot 2 \cdot a \cdot a \cdot c \end{align*}. That means when you are trying to get the lowest common denominator, in your first denominator, you are missing just \displaystyle \begin{align*} 2\cdot 2 \cdot a \cdot a \end{align*}, while in the second you are missing \displaystyle \begin{align*} 3 \cdot b \cdot c \end{align*}.

    So when you write these with the lowest common denominator

    \displaystyle \begin{align*} \frac{5a}{6bc^2} + \frac{3b}{8a^2c} &= \frac{5a \cdot 2 \cdot 2 \cdot a \cdot a}{6bc^2 \cdot 2 \cdot 2 \cdot a \cdot a} + \frac{3b \cdot 3 \cdot b \cdot c}{8a^2c \cdot 3 \cdot b \cdot c} \\ &= \frac{20a^3}{24a^2 b c^2} + \frac{9b^2 c}{24 a^2 b c^2} \end{align*}
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    Re: These Exponents Don't Work Out

    Quote Originally Posted by Prove It View Post
    First of all, this is VERY hard to read, some brackets would help. I assume that your expression is \displaystyle \begin{align*} \frac{5a}{6bc^2} + \frac{3b}{8a^2c} \end{align*}, if you look at the factors in each denominator, you have \displaystyle \begin{align*} 2 \cdot 3 \cdot b \cdot c \cdot c \end{align*}, and in the second you have \displaystyle \begin{align*} 2 \cdot 2 \cdot 2 \cdot a \cdot a \cdot c \end{align*}. That means when you are trying to get the lowest common denominator, in your first denominator, you are missing just \displaystyle \begin{align*} 2\cdot 2 \cdot a \cdot a \end{align*}, while in the second you are missing \displaystyle \begin{align*} 3 \cdot b \cdot c \end{align*}.

    So when you write these with the lowest common denominator

    \displaystyle \begin{align*} \frac{5a}{6bc^2} + \frac{3b}{8a^2c} &= \frac{5a \cdot 2 \cdot 2 \cdot a \cdot a}{6bc^2 \cdot 2 \cdot 2 \cdot a \cdot a} + \frac{3b \cdot 3 \cdot b \cdot c}{8a^2c \cdot 3 \cdot b \cdot c} \\ &= \frac{20a^3}{24a^2 b c^2} + \frac{9b^2 c}{24 a^2 b c^2} \end{align*}
    I'm trying to apply your advice to my problem, but I'm not having much luck with it... Is there an easier way about this?
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    Re: These Exponents Don't Work Out

    What do you mean by "my problem"? Prove It gave a detailed solution to the problem you gave in your first post. If you have another problem also, then completely factor the denominators and look for factors in one that are not in the other.
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    Re: These Exponents Don't Work Out

    Quote Originally Posted by MathClown View Post
    I'm trying to apply your advice to my problem, but I'm not having much luck with it... Is there an easier way about this?
    To find the \text{LCM} first factor: ~2\cdot 3\cdot b\cdot c^2~\&~2^3\cdot a^2\cdot c

    Then list each factor with its highest power: ~2^3\cdot 3\cdot a^2\cdot b\cdot c^2
    There you have it.

    Having done that it is easy to see what the numerator must be multiplied by.
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