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I was trying to solve this for two days. how can I find the value of $K$ which is the carrying capacity and the value of a, the equation shows below. $$ \frac{a(1-(26.273/K))}{a(1-(27.165/K))}=\frac{0.03274448}{0.03253040} $$

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    $\begingroup$ Is $a$ a constant? For me, it looks like it can be cancelled away in the equation. Also, please improve the mathematical notation with MathJax so that we know what the exact details in the formula are. You mentioned that you have tried to solve it - what did you try so far? What was the result? $\endgroup$ – Matti P. Feb 14 at 9:08
  • $\begingroup$ I have cancelled the constant a by ratio&proportion but i was stuck to get the value of K. $\endgroup$ – Jessabelle B. Feb 14 at 9:15
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After cancelling $a$, you are left with the equation $$\tag{1} \frac{(1-(26.273/K))}{(1-(27.165/K))}=\frac{0.03274448}{0.03253040} = 1.0065809 $$ In order to remove numerical rounding errors, let's write this as $$\tag{2} \frac{1-\frac{a}{K}}{1-\frac{b}{K}}= c $$ You can multiply both sides by $1-\frac{b}{K}$ to get $$\tag{3} 1-\frac{a}{K} = c\left(1-\frac{b}{K} \right) = c - \frac{bc}{K} $$ Let's gather terms with $K$ to the left-hand side and the terms without $K$ to the right-hand side: $$\tag{4} \frac{bc}{K} - \frac{a}{K} = c-1 $$ $$\tag{5} \frac{bc-a}{K} = c-1 $$ Continuing, you can multiply both sides by $K$ to get $$\tag{6} bc-a = (c-1)K $$ and now divide by $c-1$ to get the answer: $$\tag{7} K = \frac{bc-a}{c-1} $$ Is it clear now?

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  • $\begingroup$ i'm still confused. am i going to derive it then? or just distribute the value of the variables? $\endgroup$ – Jessabelle B. Feb 14 at 9:24
  • $\begingroup$ If there is anything unclear, please ask. I can help you. $\endgroup$ – Matti P. Feb 14 at 9:29
  • $\begingroup$ thank you so much. this is a very big help. $\endgroup$ – Jessabelle B. Feb 14 at 9:30
  • $\begingroup$ You're welcome. You can mark my answer as accepted, if you feel like it answers your question. $\endgroup$ – Matti P. Feb 14 at 9:31
  • $\begingroup$ it's done. thanks again. and I hope you will answer if i have questions later on. :) $\endgroup$ – Jessabelle B. Feb 14 at 9:33

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