Steps in finding the carrying capacity K and the value of a

Question

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. $$ \frac{a(1-(26.273/K))}{a(1-(27.165/K))}=\frac{0.03274448}{0.03253040} $$

Answer 1

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?