Simplify Fraction with same term in Num and Dem, but.

AI Thread Summary
To simplify the fraction ab/(ab+cd), factoring out common terms in the denominator is suggested. The discussion reveals that alternate forms, such as 1/(1+(cd)/(ab)), may not significantly simplify the expression. The user clarifies their context, indicating they are working on transfer functions for circuit design, which influences their desired format. The conversation concludes with a light acknowledgment of the assistance provided. Understanding the context of use is essential for determining the best form of the expression.
zKarp
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Homework Statement


I'm having a brain fart, how do you factor/simplify a fraction such as \frac{ab}{ab+cd}
I keep thinking multiply by ab-bc but I'm not sure.
 
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zKarp said:

Homework Statement


I'm having a brain fart, how do you factor/simplify a fraction such as \frac{ab}{ab+bc}
I keep thinking multiply by ab-bc but I'm not sure.

Uh, the terms in the denominator have common factor b. Why don't you factor it out?
 
Dick said:
Uh, the terms in the denominator have common factor b. Why don't you factor it out?

I'm sorry I miss typed it after using latex format. It's suppose to be DC not BC.
 
zKarp said:
I'm sorry I miss typed it after using latex format. It's suppose to be DC not BC.

Then I don't think there is any form that is terribly much simpler than what you written. You can change the form, like to 1/(1+(cd)/(ab)), but I don't think the alternate forms are much simpler.
 
Dick said:
Then I don't think there is any form that is terribly much simpler than what you written. You can change the form, like to 1/(1+(cd)/(ab)), but I don't think the alternate forms are much simpler.

Thank you! This actually does. I should of stated I'm doing transfer functions for circuit design and that's the type of form desired. Thank you!
 
zKarp said:
Thank you! This actually does. I should of stated I'm doing transfer functions for circuit design and that's the type of form desired. Thank you!

Ok, so I helped by accident. You're welcome!
 
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Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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