We can do this by combining the resistors in series and parallel.
The 12Ω and 6Ω resistors are in parallel, so their combined resistance is:
$$\frac{1}{R_{12,6}} = \frac{1}{12} + \frac{1}{6} = \frac{1}{4}$$
$$R_{12,6} = 4\,\Omega$$
Now, the combined resistance
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