robert mclaughlin
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- Jan 12, 2013
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4/(4-a^2)-5/(6+3a)
=
(2a)/(12-3a^2)
=
a/(6-3a)
i dont know if im getting the denom right
=
(2a)/(12-3a^2)
=
a/(6-3a)
i dont know if im getting the denom right
unlike denominators
- Thread starter robert mclaughlin
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- Nov 24, 2012
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- 3,021
We are given to simplify
\(\displaystyle \dfrac{4}{4-a^2}-\dfrac{5}{6+3a}\)
The first thing you want to do is factor the denominators so that you may determine the LCD:
\(\displaystyle \dfrac{4}{(2+a)(2-a)}-\dfrac{5}{3(2+a)}\)
So, what will the LCD be? Actually you have the correct LCD in your second step, but the numerator is incorrect. I think if you state the LCD in its factored form, you will find getting the numerator to be easier.
\(\displaystyle \dfrac{4}{4-a^2}-\dfrac{5}{6+3a}\)
The first thing you want to do is factor the denominators so that you may determine the LCD:
\(\displaystyle \dfrac{4}{(2+a)(2-a)}-\dfrac{5}{3(2+a)}\)
So, what will the LCD be? Actually you have the correct LCD in your second step, but the numerator is incorrect. I think if you state the LCD in its factored form, you will find getting the numerator to be easier.
robert mclaughlin
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2a/(6-3a) ??? off to school
D
Deleted member 4993
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Please show your work, step-by-step, so that we find out the mistakes you are making and we can teach you how to avoid those mistakes.
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This is a very good suggestion, Robert.
We have no idea how you came up with your incorrect result. Giving you the answer would be far less instructive than if we see what you did and point out where you went wrong, so that you can avoid this in the future.