A battery of 12.6 V and 0.050 Ω is joined in parallel with an older battery of 11.4 V and 0.080 Ω…

Step 1. The top line. $12.6\ \text{V} \times 0.080\ \Omega = 1.008\ \text{V}\,\Omega$ and $11.4\ \text{V} \times 0.050\ \Omega = 0.570\ \text{V}\,\Omega$. Together they make $1.578\ \text{V}\,\Omega$.

Step 2. The pair. $E = \dfrac{1.578\ \text{V}\,\Omega}{0.130\ \Omega} = 12.14\ \text{V}$. The internal resistance is $r = \dfrac{0.050 \times 0.080}{0.130}\ \Omega = 0.0308\ \Omega$.

Step 3. The motor. $I = \dfrac{12.14\ \text{V}}{0.20\ \Omega + 0.0308\ \Omega} = 52.6\ \text{A}$. The voltage is $V = 52.6\ \text{A} \times 0.20\ \Omega = 10.52\ \text{V}$.

Step 4. Each battery. The first gives $\dfrac{12.6\ \text{V} - 10.52\ \text{V}}{0.050\ \Omega} = 41.6\ \text{A}$. The second gives $\dfrac{11.4\ \text{V} - 10.52\ \text{V}}{0.080\ \Omega} = 11.0\ \text{A}$.