cjquines0 wrote: Let $\mathbb{R}^*$ be the set of all real numbers, except $1$. Find all functions $f:\mathbb{R}^* \rightarrow \mathbb{R}$ that satisfy the functional equation $$x+f(x)+2f(\frac{x+2009}{x-1})=2010$$. I suppose that domain of functional equaiton is the same as domain of function : $\mathbb R\setminus\{1\}$ Is so : $x+f(x)+2f(\frac{x+2009}{x-1})=2010$ Setting there $x\to\frac{x+2009}{x-1}$, this gives $\frac{x+2009}{x-1}+f(\frac{x+2009}{x-1})+2f(x)=2010$ Cancelling $f(\frac{x+2009}{x-1})$ between these two lines, we get $\boxed{f(x)=\frac 13\left(x+2010-2\frac{x+2009}{x-1}\right)}$ which indeed is a solution