How can 1/y = kr + C/e^(rt) be solved for y?

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Homework Statement



1/y = kr + C/e^(rt)

Is this equal to y = 1/kr + e^(rt)/C

Thanks


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The Attempt at a Solution

 
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No, 1/a = b + c is not a = 1/b + 1/c but rather a = 1/(b+c)

Think about it this way 3 = 2 + 1 but would you say that 1/3 = 1/2 + 1/1?
 
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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