I labeled it up like this. It's about the same. I just dumped the \$+6\:\text{V}\$ 2-terminal voltage source device as unneeded:
simulate this circuit – Schematic created using CircuitLab
The equations are then:
$$\begin{align*}
I_{\small{C}}&=\frac{V_{\small{C}}}{R_5}
\\\\
\frac{V_{\small{B}}}{R_1}+\frac{V_{\small{B}}}{R_2}&=\frac{V_{\small{E}}}{R_2}+\frac12 I_{\small{C}}
\\\\
\frac{V_{\small{E}}}{R_2}+\frac{V_{\small{E}}}{R_3}+I_1&=\frac{V_{\small{B}}}{R_2}+\frac{V_{\small{D}}}{R_3}+2\:\text{S}\cdot \left(V_{\small{A}}-V_{\small{C}}\right)
\\\\
\frac{V_{\small{D}}}{R_3}+\frac12 I_{\small{C}}&=\frac{V_{\small{E}}}{R_3}+I_x
\\\\
\frac{V_{\small{A}}}{R_4}+I_x+500\:\text{mS}\cdot V_{\small{B}}&=\frac{V_{\small{C}}}{R_4}
\\\\
\frac{V_{\small{C}}}{R_4}+\frac{V_{\small{C}}}{R_5}&=\frac{V_{\small{A}}}{R_4}+I_1+I_2
\\\\
V_{\small{A}}&=V_{\small{D}}+V_1
\end{align*}$$
That solves out as
va: 18
vb: 4
vc: 16
vd: -6
ve: 3
ix: -2.5
An LTspice run confirms the values:
The matrix form of the above would be:
$$
\!\!\!\!\!\!\!\!\!\!{\begin{align*}&V_A & \!V_B && V_C&&V_D&&\!\!V_E&&\!\!I_x\end{align*}}
\\
\begin{matrix}\text{KCL }V_A\vphantom{\frac{-1}2}\\\text{KCL }V_B\vphantom{\frac{-1}2}\\\text{KCL }V_C\vphantom{\frac{-1}2}\\\text{KCL }V_D\vphantom{\frac{-1}2}\\\text{KCL }V_E\vphantom{\frac{-1}2}\\\text{KVL }V_A-V_D\\\end{matrix}
\begin{bmatrix}
\frac14&\frac12&\frac{-1}4&0&0&1
\\
0&\frac11+\frac14&\frac{-1}{2\,\cdot\, 4}&0&\frac{-1}1&0
\\
\frac{-1}4&0&\frac14+\frac14&0&0&0
\\
0&0&\frac1{2\,\cdot\, 4}&\frac12&\frac{-1}2&-1
\\
-2&\frac{-1}1&2&\frac{-1}2&\frac11+\frac12&0
\\
1&0&0&-1&0&0
\end{bmatrix}
\cdot
\begin{bmatrix}V_A\vphantom{\frac{-1}2}\\V_B\vphantom{\frac{-1}2}\\V_C\vphantom{\frac{-1}2}\\V_D\vphantom{\frac{-1}2}\\V_E\vphantom{\frac{-1}2}\\I_x\end{bmatrix}
=
\begin{bmatrix}0\vphantom{\frac{-1}2}\\0\vphantom{\frac{-1}2}\\3.5\vphantom{\frac{-1}2}\\0\vphantom{\frac{-1}2}\\-0.5\vphantom{\frac{-1}2}\\24\end{bmatrix}$$
Using SymPy/Python/SageMath I get:
A = Matrix( [
[ 1/4, 1/2, -1/4, 0, 0, 1 ],
[ 0, 1/1+1/4, -1/(2*4), 0, -1/1, 0 ],
[ -1/4, 0, 1/4+1/4, 0, 0, 0 ],
[ 0, 0, 1/(2*4), 1/2, -1/2, -1 ],
[ -2, -1/1, 2, -1/2, 1/1+1/2, 0 ],
[ 1, 0, 0, -1, 0, 0 ] ] )
A
Matrix([
[ 1/4, 1/2, -1/4, 0, 0, 1],
[ 0, 5/4, -1/8, 0, -1, 0],
[-1/4, 0, 1/2, 0, 0, 0],
[ 0, 0, 1/8, 1/2, -1/2, -1],
[ -2, -1, 2, -1/2, 3/2, 0],
[ 1, 0, 0, -1, 0, 0]])
A.inv()*Matrix([[0],[0],[3.5],[0],[-0.5],[24]])
Matrix([
[18.0],
[ 4.0],
[16.0],
[-6.0],
[ 3.0],
[-2.5]])
Which is the same result as before.
