I have the following equation, which i am supposed to solve for $s_{0,4}$: $$P=\frac{C}{1+s_{0,1}}+\frac{C}{(1+s_{0,2})^2}+\frac{C}{(1+s_{0,3})^3}+\frac{C+F}{(1+s_{0,4})^4}$$ which I rearranged in the following steps : 1. $$ P-\frac{C}{1+s_{0,1}}-\frac{C}{(1+s_{0,2})^2}-\frac{C}{(1+s_{0,3})^3}=\frac{C+F}{(1+s_{0.4})^4} $$ 2. $$ \frac{P-\frac{C}{1+s_{0,1}}-\frac{C}{(1+s_{0,2})^2}-\frac{C}{(1+s_{0,3})^3}}{C+F}=\frac{1}{(1+s_{0.4})^4} $$ My question is if could just take the reciprocal of both sides and turn the equation into this $$\frac{C+F}{P-\frac{C}{1+s_{0,1}}-\frac{C}{(1+s_{0,2})^2}-\frac{C}{(1+s_{0,3})^3}}=(1+s_{0.4})^4$$ which would then be easily solvable for $s_{0,4}$
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$\begingroup$ What happens when you try to prove it? Assume $A=B$. Deduce that $1/A = 1/B$. $\endgroup$– GEdgarCommented May 24 at 19:22
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$\begingroup$ That's not what you're doing though - you're pushing the $\frac{C}{1+s_{0,k}}$ terms to the LHS first, and then taking the reciprocals. $\endgroup$– Sharat V ChandrasekharCommented May 24 at 21:06
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