Finding Thevenin equivalent
Let's assume there are two parallel voltage sources \$V_1\$ and \$V_2\$ with internal resistances \$R_1\$ and \$R_2\$:



To compute the Thevenin equivalent you need the open-circuit voltage \$V_{TH}\$ and the total internal resistance \$R_{Th}\$, or, equivalently, the short-circuit current \$I_{Th}\$. For \$V_{Th}\$ you obtain
$$V_{Th} = \frac{V_1R_2+V_2R_1}{R_1+R_2}$$

This is most easily seen by using the superposition principle, i.e. by setting \$V_2=0\$ and computing the contribution of \$V_1\$ and then doing the same for \$V_2\$.
For \$I_{Th}\$ we get
$$I_{Th} = \frac{V_1}{R_1} + \frac{V_2}{R_2}$$
\$R_{Th}\$ is then given by
$$R_{Th} = \frac{V_{Th}}{I_{Th}} = \frac{R_1R_2}{R_1+R_2}$$
The total internal resistance \$R_{Th}\$ can also be directly obtained by noting that it is the parallel resistance of \$R_1\$ and \$R_2\$.