How Can Complex Fractions be Simplified?

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The discussion focuses on simplifying complex fractions, specifically the expression (a - b)/(a-1 - b-1). Participants emphasize the importance of recognizing negative exponents and the need for a common denominator when subtracting fractions. A step-by-step method is provided, including inverting fractions and using the least common denominator (LCD) to simplify the expression correctly. The final simplified result is confirmed as -ab. The forum is praised for its supportive environment, helping users improve their math skills and confidence.
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First of all, thank you, everyone, for all your help with my math conundrums! I've really appreciated the patience and helpful hints - because I do want to learn it myself, so this method is just enough for me to cut the math problems down to size. I'm thankful. And, I have noticed my scores improved because I'm recognizing my trouble areas more often. So, to anyone out there shy to ask for help, I do suggest this forum. It's been very useful!

Homework Statement


Simplify. (a - b)/(a-1 - b-1)

Answer: -ab

This lesson is teaching that you can either simplify the numerator and denominator separately and divide; or multiply the whole thing by the LCD and then simpify.

Homework Equations


--

The Attempt at a Solution


(a - b)/(a-1 - b-1)
Switched to a different format.
(a-b)/1 ÷ (a-1 - b-1)/1
Get rid of negative exponents.
(a-b)/1 ÷ 1/(a - b)
Invert the second fraction & switch to multiplication.
(a-b)/1 X (a-b)/1
Perform multiplication.
a2 - 2ab - b2
 
Last edited:
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velox_xox said:
(a-b)/1 ÷ (a-1 - b-1)/1
Get rid of negative exponents.
(a-b)/1 ÷ 1/(a - b)

Hi velox! This step is wrong! It might be easier to see the two cases you are confusing in fraction form: a^{-1}-b^{-1}= \displaystyle \frac{1}{a}-\frac{1}{b} and (a-b)^{-1} = \displaystyle \frac{1}{a-b}

Whatever is being raised to the negative one gets flipped. All nearby operators stay the same. So in the first case that I showed above, the a and the b, individually, are raised to the negative one. So a and b, individually, get "flipped" and the minus sign stays where it is. In the second case the whole expression is raised to the negative one; so the whole expression gets flipped.

I hope that helps! Try getting the correct answer now!
 
Hi scurty, it's nice to have your help again!

That's right, the denominators need to be the same in subtraction with fractions, so I see that the separate 1/a - 1/b and 1/a-b aren't the same. I guess I need to watch out for that.

So, then...
(a - b)/(1/a - 1/b)
Use the LCD of 'ab' to create a common denominator.
{(a - b)/(b - a)/(ab)}
Turn the complex fraction into two simple ones.
(a - b)/1 ÷ (b - a)/(ab)
Invert the second fraction and change the sign to multiplication (and also change the denominator of the second fraction to the proper form).
(a - b)/1 X (ab)/(-a + b)
Factor out the '-1' from the denominator of the second fraction.
(a - b)/1 X (ab)/-1(a - b)
Simplify.
(ab)/-1
Divide.
-ab

Is this correct??
 
Absolutely! I see no problem with your method!
 
Well, that was fast. Thank you so much scurty! Have a good weekend!
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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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