If the conditions 
\[\begin{array}{rcl} 
f(2x+1)+g(3-x) &=& x \\
f((3x+5)/(x+1))+2g((2x+1)/(x+1)) &=& x/(x+1)
\end{array}\] hold for all real numbers $x\neq 1$, what is $f(2013)$?


$ 
\textbf{(A)}\ 1007
\qquad\textbf{(B)}\ \dfrac {4021}{3}
\qquad\textbf{(C)}\ \dfrac {6037}7
\qquad\textbf{(D)}\ \dfrac {4029}{5}
\qquad\textbf{(E)}\ \text{None of above}
$
1. We start with the given equations:
   \[
   f(2x+1) + g(3-x) = x
   \]
   \[
   f\left(\frac{3x+5}{x+1}\right) + 2g\left(\frac{2x+1}{x+1}\right) = \frac{x}{x+1}
   \]

2. Let \( P(x) \) be the assertion of the first equation:
   \[
   P(x): f(2x+1) + g(3-x) = x
   \]

3. Substitute \( x = \frac{y+2}{y+1} \) into \( P(x) \):
   \[
   P\left(\frac{y+2}{y+1}\right): f\left(\frac{3y+5}{y+1}\right) + g\left(\frac{2y+1}{y+1}\right) = \frac{y+2}{y+1}
   \]

4. Now, we have the second given equation:
   \[
   f\left(\frac{3x+5}{x+1}\right) + 2g\left(\frac{2x+1}{x+1}\right) = \frac{x}{x+1}
   \]

5. Subtract the second given equation from twice \( P\left(\frac{x+2}{x+1}\right) \):
   \[
   2 \left( f\left(\frac{3x+5}{x+1}\right) + g\left(\frac{2x+1}{x+1}\right) \right) - \left( f\left(\frac{3x+5}{x+1}\right) + 2g\left(\frac{2x+1}{x+1}\right) \right) = 2 \cdot \frac{x+2}{x+1} - \frac{x}{x+1}
   \]
   Simplifying the left-hand side:
   \[
   f\left(\frac{3x+5}{x+1}\right) + 2g\left(\frac{2x+1}{x+1}\right) - f\left(\frac{3x+5}{x+1}\right) - 2g\left(\frac{2x+1}{x+1}\right) = \frac{2(x+2) - x}{x+1}
   \]
   \[
   f\left(\frac{3x+5}{x+1}\right) = \frac{x+4}{x+1}
   \]

6. To find \( f(2013) \), we need to find \( x \) such that \( 2x+1 = 2013 \):
   \[
   2x + 1 = 2013
   \]
   \[
   2x = 2012
   \]
   \[
   x = 1006
   \]

7. Substitute \( x = 1006 \) into the expression for \( f \):
   \[
   f(2013) = \frac{1006 + 4}{1006 + 1} = \frac{1010}{1007}
   \]

The final answer is \(\boxed{\frac{1010}{1007}}\)