What is the voltage drop across the 10 ohm resistor?
Millman's theorem is a kind of multi source voltage divider theorem based on this topology: -


So, re-arrange your circuit to use a 3rd source that is actually 0 volts and you're good to go: -

Hence \$V_{AB} = \dfrac{\frac{10}{3.33333} + \frac{0}{10} +\frac{20}{4}}{\frac{1}{3.3333}+ \frac{1}{10} + \frac{1}{4}} = \dfrac{8}{0.65} = 12.3077\$
But equally, without working out the parallel resistance of the 10 Ω and 5 Ω you could have written this (now there are two sources of 10 volts, one in series with 10 Ω and one in series with 5 Ω): -
$$V_{AB} = \dfrac{\frac{10}{10} +\frac{10}{5} + \frac{0}{10} +\frac{20}{4}}{\frac{1}{10}+\frac{1}{5}+ \frac{1}{10} + \frac{1}{4}} = \dfrac{8}{0.65} = 12.3077$$