Biasing in a radio receiver circuit

An answer:

\begin{array}{rcl} I_{C} & = \frac{βfe}{βfe + 1} I_{E} \\ I_{E} = \frac{V_{E}}{R_{6}} \\ V_{E} = V_{B} - V_{BE} \\ V_{B} = V_{CC} \cdot \frac{R_{4} + R_{5}}{R_{3} + R_{4} + R_{5}} - I_{B} \cdot \frac{(R_{4} + R_{5}) \cdot R_{3}}{R_{3} + R_{4} + R_{5}} \\ ⟹ \\ I_{C} & = \frac{βfe}{βfe + 1} \frac{V_{CC} \cdot \frac{R_{4} + R_{5}}{R_{3} + R_{4} + R_{5}} - 0.6 V}{R_{6} + \frac{1}{βfe + 1} \cdot \frac{(R_{4} + R_{5}) \cdot R_{3}}{R_{3} + R_{4} + R_{5}}} \end{array}

In this question, letting $βfe \to \infty$ from the start:

\begin{array}{rcl} I_{C} & = I_{E} \\ I_{E} = \frac{V_{E}}{R_{6}} \\ V_{E} = V_{B} - V_{BE} \\ V_{B} = V_{CC} \cdot \frac{R_{4} + R_{5}}{R_{3} + R_{4} + R_{5}} \\ ⟹ \\ I_{C} & = \frac{V_{CC} \cdot \frac{R_{4} + R_{5}}{R_{3} + R_{4} + R_{5}} - 0.6 V}{R_{6}} \end{array}