Solve for the unknown node voltage V_X in the given circuit diagram.

We choose the bottom wire as the reference node, so its voltage is $0\,\text{V}$.
There is one unknown node voltage, which we will call $V_X$. This is the top node where resistors $R_1$, $R_2$, and $R_3$ connect.
The voltage at the left end of $R_1$ is $28\,\text{V}$ (due to battery $B_1$).
The voltage at the right end of $R_3$ is $7\,\text{V}$ (due to battery $B_2$).

According to KCL, the sum of currents entering a node must be equal to the sum of currents leaving the node. From the diagram, currents $I_1$ and $I_3$ are entering node $V_X$, and current $I_2$ is leaving node $V_X$.

Now, express these currents using Ohm's Law ($I = \frac{V}{R}$):

• $I_1$ flows from $28\,\text{V}$ to $V_X$ through $R_1 = 4\,\Omega$: $$I_1 = \frac{28\,\text{V} - V_X}{4\,\Omega}$$
• $I_2$ flows from $V_X$ to $0\,\text{V}$ through $R_2 = 2\,\Omega$: $$I_2 = \frac{V_X - 0\,\text{V}}{2\,\Omega} = \frac{V_X}{2\,\Omega}$$
• $I_3$ flows from $7\,\text{V}$ to $V_X$ through $R_3 = 1\,\Omega$: $$I_3 = \frac{7\,\text{V} - V_X}{1\,\Omega}$$

Substitute these expressions into the KCL equation:

To eliminate the denominators, multiply the entire equation by the least common multiple of 4, 1, and 2, which is 4:

Distribute the 4:

Combine like terms:

Add $5V_X$ to both sides:

Divide by 7:

The voltage at the top node is $8\,\text{V}$.