# These Exponents Don't Work Out

• Oct 18th 2012, 06:21 AM
MathClown
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.
• Oct 18th 2012, 06:30 AM
Prove It
Re: These Exponents Don't Work Out
Quote:

Originally Posted by MathClown
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*}
• Oct 18th 2012, 07:05 AM
MathClown
Re: These Exponents Don't Work Out
Quote:

Originally Posted by Prove It
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?
• Oct 18th 2012, 07:39 AM
HallsofIvy
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.
• Oct 18th 2012, 07:52 AM
Plato
Re: These Exponents Don't Work Out
Quote:

Originally Posted by MathClown
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.