# Not so basic algebraic fraction manipulation

by namtip
Tags: algebraic, basic, fraction, manipulation
 P: 5 1. The problem statement, all variables and given/known data I am told this is wrong, but I am convinced it is right. I just need to show how but don't have a clue. 2. Relevant equations (ab)/(2c(d+e))+(ab)/(2c(d-e))=(abd)/(c(d^2-e^2)) 3. The attempt at a solution I do not have a clue where to start. I am clueless as to where the d in the numerator comes from... Any help at all, even a nudge would be very much appreciated. Thanks
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P: 1,965
 Quote by namtip 1. The problem statement, all variables and given/known data I am told this is wrong, but I am convinced it is right. I just need to show how but don't have a clue. 2. Relevant equations (ab)/(2c(d+e))+(ab)/(2c(d-e))=(abd)/(c(d^2-e^2)) 3. The attempt at a solution I do not have a clue where to start. I am clueless as to where the d in the numerator comes from... Any help at all, even a nudge would be very much appreciated. Thanks
Didn't take me long but I don't make a virtue of that - practice does help.

You have no idea where to start?!

Come on - I don't believe you have never done anything like it.

Start making it easier by taking out a factor common that multiplies both parts of the expression.
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P: 7,799
 Quote by namtip 1. The problem statement, all variables and given/known data I am told this is wrong, but I am convinced it is right. I just need to show how but don't have a clue. 2. Relevant equations (ab)/(2c(d+e))+(ab)/(2c(d-e))=(abd)/(c(d^2-e^2)) 3. The attempt at a solution I do not have a clue where to start. I am clueless as to where the d in the numerator comes from... Any help at all, even a nudge would be very much appreciated. Thanks
I'm just converting those expressions to LaTeX so they're easier to read.

$\displaystyle \frac{ab}{2c(d+e)}+\frac{ab}{2c(d-e)}\text{ is possibly equal to }\frac{abd}{c(d^2-e^2)}$

What makes you think they're equal. If you're sure that they are equal, can you show the steps you take to get from the left hand expression to the right hand expression.

Also, who told you that you're wrong?

 P: 5 Not so basic algebraic fraction manipulation Seriously, my algebra is terrible. So I take out a factor that is common to both sides? I take it you mean both sides of the equals sign. That would be the 'ab' then. But when you say 'take out' I'm not sure what you mean. Like this? ab*1/(2c(d+e))+ab*1/(2c(d-e))=ab*d/(c(d^2-e^2))
 P: 5 Hi SammyS I wrote that first sentence wrong, I meant 'I am told this is right, but I think its wrong'. See? If I can't put a sentence together what hope do I have in algebra?! This is an example I got given and need to show the processes. I've solved all the others given to me except this one. It was thrown in as an 'extra hard challenge'.
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P: 7,799
 Quote by namtip Hi SammyS I wrote that first sentence wrong, I meant 'I am told this is right, but I think its wrong'. See? If I can't put a sentence together what hope do I have in algebra?! This is an example I got given and need to show the processes. I've solved all the others given to me except this one. It was thrown in as an 'extra hard challenge'.
Well ... It is absolutely correct.

Find a common denominator (the 'smallest' common denominator is best.) and use it to add the fractions.
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Thanks
P: 4,955
 Quote by namtip Seriously, my algebra is terrible. So I take out a factor that is common to both sides? I take it you mean both sides of the equals sign. That would be the 'ab' then. But when you say 'take out' I'm not sure what you mean. Like this? ab*1/(2c(d+e))+ab*1/(2c(d-e))=ab*d/(c(d^2-e^2))
No; taking out a common factor means recognizing that
$$\frac{ab}{2c(d+e)} + \frac{ab}{2c(d-e)}$$
can be written as
$$\frac{ab}{2c} \left[ \frac{1}{d+e} + \frac{1}{d-e} \right],$$
so it is enough to simplify the part in square brackets; that is, it is enough to figure out what
$$\frac{1}{d+e} + \frac{1}{d-e}$$
simplifies to. After that you can put back the factor $ab/(2c).$

Note: typo fixed.

RGV
 HW Helper P: 1,965 You meant $$\frac{ab}{2c}$$ (else may confuse).
 Quote by epenguin You meant $$\frac{ab}{2c}$$ (else may confuse).