Identify the series combination of the 3 and 8 resistors.

The equivalent resistance for resistors in series is the sum of their resistances.

$$R_{series1} = 3 \, \Omega + 8 \, \Omega = 11 \, \Omega$$

The equivalent resistance for resistors in parallel is given by $\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2}$.

$$\frac{1}{R_{parallel}} = \frac{1}{6 \, \Omega} + \frac{1}{11 \, \Omega}$$ $$\frac{1}{R_{parallel}} = \frac{11}{66 \, \Omega} + \frac{6}{66 \, \Omega}$$ $$\frac{1}{R_{parallel}} = \frac{11 + 6}{66 \, \Omega}$$ $$\frac{1}{R_{parallel}} = \frac{17}{66 \, \Omega}$$ $$R_{parallel} = \frac{66}{17} \, \Omega$$

The total equivalent resistance $R_1$ is the sum of these resistances.

$$R_1 = 2 \, \Omega + \frac{66}{17} \, \Omega$$ $$R_1 = \frac{2 \times 17}{17} \, \Omega + \frac{66}{17} \, \Omega$$ $$R_1 = \frac{34}{17} \, \Omega + \frac{66}{17} \, \Omega$$ $$R_1 = \frac{34 + 66}{17} \, \Omega$$ $$R_1 = \frac{100}{17} \, \Omega$$

The equivalent resistance $R_1$ is:

$$\frac{100}{17} \, \Omega$$