jmc

We begin with a system of two equations $$\begin{array}{rl}2d& =17e-8\\ 2e& =d-9\end{array}$$Since the second equation can also be rewritten as $d=2e+9$, we can plug this expression for $d$ back into the first equation and solve for $e$ $$\begin{array}{rl}2d& =17e-8\\ \Rightarrow 2(2e+9)& =17e-8\\ \Rightarrow 4e+18& =17e-8\\ \Rightarrow -13e& =-26\\ \Rightarrow e& =\boxed{2}.\end{array}$$