Biasing an amplifier with an NPN

An answer:

\begin{array}{rcl} I_{C} = & \frac{βfe}{βfe + 1} I_{E} \\ I_{E} = \frac{V_{E}}{R_{4}} \\ V_{E} = V_{DD} - I_{B} R_{1} - V_{BE} \\ I_{B} = \frac{I_{C}}{βfe} \end{array}

back substitution and separation of variables yield

\begin{array}{rcl} I_{C} = & \frac{βfe}{βfe + 1} \frac{V_{DD} - \frac{I_{C}}{βfe} R_{1} - V_{BE}}{R_{4}} \\ = \frac{βfe}{βfe + 1} \frac{V_{DD} - V_{BE}}{R_{4} + \frac{R_{1}}{βfe + 1}} \end{array}