Real World Applications of the (x-y)/(x+y) Equation

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The discussion focuses on the real-world applications of the (x-y)/(x+y) equation, particularly in understanding ratios and differences. The original poster shares a scenario involving selling oranges and expresses confusion about the application of the equation, specifically how it yields a 60% result. Another participant clarifies that (x-y)/(x+y) can indicate balance in groups, using a party scenario with girls and boys as an example. The conversation highlights the equation's utility in analyzing proportions and differences in various contexts. Overall, the equation serves as a valuable tool for understanding ratios in real-life situations.
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Homework Statement


Real world use of (x-y)/(x+y).


Homework Equations


(x-y)/x


The Attempt at a Solution


(x-y)/x solves for the mean difference ratio. I had 400 oranges and sold 100. 400-100/400 gives .75 or 75% remaining. I do not understand where I would use the difference over sum equation. The answer: .6 or 60% using this equation baffles me.

Can anyone set me straight on this.

Thanks!
 
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hi mathisp64! welcome to pf! :smile:
mathisp64 said:
(x-y)/x solves for the mean difference ratio. I had 400 oranges and sold 100. 400-100/400 gives .75 or 75% remaining. I do not understand where I would use the difference over sum equation. The answer: .6 or 60% using this equation baffles me.

you have a party with 70 girls and 62 boys …

(x-y)/(x+y) tells you how unbalanced it is :wink:
 
Thanks, that was exactly what I was needing to know.
 
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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