Determine the current through a voltage source for nodal analysis
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ANSWER: We don't need to use the third equation, actually. We substitute the current representations for \$I_{AC}, I_{GND-A}, I_{GND-B}, I_{AB}\$ and \$I_{CB}\$ in the respective equations for the nodes A and B. Which means:

Node A: \$-\frac{E_A-2}{3}+\frac{E_A}{2}+1=0\$

Node B: \$-1-E_B+\frac{2-E_B}{x}=0\$  (\$E_C = 2V\$) 

Solving the equation for node A we get \$E_A\$, and solving the equation of node B we get \$E_B\$. The potential difference between the node B and the ground (V_1) is equal to \$E_B\$, which is a function of x. Therefore:

\$V_1=\frac{2-x}{x+1}\$, and because \$R_1 = 1Ω, I_1=V_1\$.

Finally, we get that: \$P_1 = (\frac{2-x}{x+1})^2\$ and now we need to find the value of x which maps to the maximum of this function for \$x &gt;= 0\$, which is trivial if you know a bit of Calculus I.  (Spoiler: it's \$x = 0\$)