1. **Problem statement:** Find the current $I_{A\to B}$ flowing from node A to node B in the given circuit with two loops, resistors of 3 $\Omega$ each, and batteries of 3 V and 9 V. 2. **Label currents:** Let $I_1$ be the current in the left loop (through the 3 V battery and left 3 $\Omega$ resistor), $I_2$ be the current in the right loop (through the 9 V battery and right 3 $\Omega$ resistor). The current from A to B through the middle 3 $\Omega$ resistor is $I_{A\to B} = I_1 - I_2$. 3. **Write loop equations using Kirchhoff's Voltage Law (KVL):** - Left loop: $$3 - 3I_1 - 3(I_1 - I_2) = 0$$ - Right loop: $$9 - 3I_2 - 3(I_2 - I_1) = 0$$ 4. **Simplify left loop equation:** $$3 - 3I_1 - 3I_1 + 3I_2 = 0$$ $$3 - 6I_1 + 3I_2 = 0$$ 5. **Simplify right loop equation:** $$9 - 3I_2 - 3I_2 + 3I_1 = 0$$ $$9 - 6I_2 + 3I_1 = 0$$ 6. **Rewrite system:** $$3 - 6I_1 + 3I_2 = 0$$ $$9 - 6I_2 + 3I_1 = 0$$ 7. **Rearranged:** $$-6I_1 + 3I_2 = -3$$ $$3I_1 - 6I_2 = -9$$ 8. **Multiply first equation by 2 to align coefficients:** $$-12I_1 + 6I_2 = -6$$ $$3I_1 - 6I_2 = -9$$ 9. **Add equations:** $$(-12I_1 + 6I_2) + (3I_1 - 6I_2) = -6 + (-9)$$ $$-9I_1 = -15$$ 10. **Solve for $I_1$:** $$I_1 = \frac{-15}{-9} = \frac{15}{9} = \frac{5}{3}$$ 11. **Substitute $I_1$ into first equation:** $$-6 \times \frac{5}{3} + 3I_2 = -3$$ $$-10 + 3I_2 = -3$$ $$3I_2 = 7$$ $$I_2 = \frac{7}{3}$$ 12. **Calculate current from A to B:** $$I_{A\to B} = I_1 - I_2 = \frac{5}{3} - \frac{7}{3} = -\frac{2}{3}$$ **Answer:** $$I_{A\to B} = -\frac{2}{3} \text{ A}$$ The negative sign indicates the current flows opposite to the assumed direction from A to B.