quiescent point of two transistors BJTs
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simulate this circuit &ndash; Schematic created using CircuitLab

You can write these KVL equations:

$$V_E = V_{BE1}+I_{B1}*R_F$$

$$V_E = (I_{E2}-I_{B1})*R_E$$

And this give us 

$$(I_{E2}-I_{B1})*R_E = V_{BE1}+I_{B1}*R_F $$

Additional we know: 

$$I_{B1}=\frac{I_{C1}}{\beta1}$$ 

we end up with this: 

$$(I_{E2}-\frac{I_{C1}}{\beta1})*R_E = V_{BE1}+\frac{I_{C1}}{\beta1}*R_F \;\;(1) $$ 

And another KVL equation is: 

$$V_{CC} - I_{RC1}*R_{C1} - V_{BE2} = V_E$$ 

where : 

$$I_{RC1} = I_{C1}+I_{B2} = I_{C1}+\frac{I_{C2}}{\beta2}$$

Hence : 

$$V_{CC} - \left ( I_{C1}+\frac{I_{C2}}{\beta2} \right)*R_{C1} - V_{BE2} = \left (I_{E2}-\frac{I_{C1}}{\beta1}\right)*R_E \;\;(2) $$ 

So, we have two equations and two unknowns. Because Q2 emitter current is:   

$$I_{E2} = I_{C2}*\frac{\beta2 +1}{\beta2} $$  

And if we assume \$\beta = \infty \$

We will end up the this: 

$$I_{C1} = \frac{V_{CC} - 2V_{BE}}{R_{C1}}$$ 
$$I_{C2} = \frac{V_{BE}}{R_E}$$