Super Node

There is no impedance associated with the middle voltage source, so we can't define a current through that source the way we usually do with nodal analysis. Instead, treat the highlighted region as a "super node" and account for the fact that the sum of the currents flowing out of it must be zero.

$(V - 5) + V + (V + 3) + (V + 3 - 10) = 0 \\ 4 \, V = 9 \\ V = 2.25$

It is also possible to solve this using current meshes, as shown below. It is important to note that some of the resistors (the "shunt" resistors) have currents which are superpositions of the defined loop currents.

KVL equations:

$5 - I_1 - (I_1 - I_2) = 0 \\ -(I_2 - I_1) + 3 - (I_2 - I_3) = 0 \\ -(I_3 - I_2) - I_3 - 10 = 0$

This yields the following:

$I_1 = 2.75 \\ I_2 = 0.5 \\ I_3 = -4.75$

And the node voltage of interest is:

$V = 5 - I_1 = 5 - 2.75 = 2.25 \\ V = 10 + I_3 - 3 = 10 - 4.75 - 3 = 2.25$

We can solve the problem using superposition theorem, by short-circuiting two voltage sources and considering only one voltage source at a time and then add the three voltages at the node at the end. Referring to the figure:

$\begin{cases} V_1 = \dfrac {1||1||1}{1+1||1||1} \times 5 = \dfrac {\frac 13}{1+\frac 13} \times 5 = \dfrac 54 \text{ V} \\ V_2 = \dfrac {1||1||1}{1+1||1||1} \times 10 = \dfrac {\frac 13}{1+\frac 13} \times 10 = \dfrac 52 \text{ V} \\ V_2 = \dfrac {1||1}{1||1+1||1} \times (-3) = \dfrac {\frac 12}{\frac 12+\frac 12} \times (-3) = - \dfrac 32 \text{ V} \end{cases}$

Then, the voltage at the node $V_{node} = V_1+V_2+V_3 = \dfrac 54 + \dfrac 52 - \dfrac 32 = \dfrac 94 = \boxed{2.25} \text{ V}$ .