Not so basic algebraic fraction manipulation

[namtip]

Homework Statement

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.

Homework Equations

(ab)/(2c(d+e))+(ab)/(2c(d-e))=(abd)/(c(d^2-e^2))

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

 

[epenguin]

Homework Statement

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.

Homework Equations

(ab)/(2c(d+e))+(ab)/(2c(d-e))=(abd)/(c(d^2-e^2))

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.

 

[SammyS]

Homework Statement

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.

Homework Equations

(ab)/(2c(d+e))+(ab)/(2c(d-e))=(abd)/(c(d^2-e^2))

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?

 

[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))

 

[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'.

 

[SammyS]

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.

 

[Ray Vickson]

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

 

[epenguin]

You meant
$$\frac{ab}{2c}$$

(else may confuse).

 

[Ray Vickson]

You meant
$$\frac{ab}{2c}$$

(else may confuse).

Right: it is fixed.

RGV