Calculating Thevenin's open circuit voltage
Well, I am trying to analyze the following circuit:



simulate this circuit &ndash; Schematic created using CircuitLab
When we use and apply KCL, we can write the following set of equations:
$$
\begin{cases}
\text{I}_\text{a}=\text{I}_1+\text{I}_5\\
\\
\text{I}_5=\text{I}_2+\text{I}_3\\
\\
\text{I}_4=\text{I}_\text{b}+\text{I}_3\\
\\
\text{I}_1=\text{I}_\text{a}+\text{I}_6\\
\\
0=\text{I}_2+\text{I}_6+\text{I}_7\\
\\
\text{I}_4=\text{I}_\text{b}+\text{I}_7
\end{cases}\tag1
$$
When we use and apply Ohm's law, we can write the following set of equations:
$$
\begin{cases}
\text{I}_1=\frac{\text{V}_1}{\text{R}_1}\\
\\
\text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_2}{\text{R}_4}\\
\\
\text{I}_\text{b}=\frac{\text{V}_3-\text{V}_2}{\text{R}_5}
\end{cases}\tag2
$$
Because you already have the answer I will present a method that uses Mathematica to solve this problem. I used the following code to solve the two systems from above:
In[1]:=FullSimplify[
 Solve[{Ia == I1 + I5, I5 == I2 + I3, I4 == Ib + I3, I1 == Ia + I6, 
   0 == I2 + I6 + I7, I4 == Ib + I7, I1 == V1/R1, I2 == V1/R2, 
   I3 == (V1 - V2)/R3, I4 == V2/R4, Ib == (V3 - V2)/R5}, {I1, I2, I3, 
   I4, I5, I6, I7, V1, V2, V3}]]

Out[1]={{I1 -&gt; (R2 (Ib R4 + Ia (R3 + R4)))/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I2 -&gt; (R1 (Ib R4 + Ia (R3 + R4)))/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I3 -&gt; (Ia R1 R2 - Ib (R1 + R2) R4)/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I4 -&gt; ((Ia + Ib) R1 R2 + Ib (R1 + R2) R3)/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I5 -&gt; (-Ib R2 R4 + Ia R1 (R2 + R3 + R4))/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I6 -&gt; (Ib R2 R4 - Ia R1 (R2 + R3 + R4))/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  I7 -&gt; (Ia R1 R2 - Ib (R1 + R2) R4)/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  V1 -&gt; (R1 R2 (Ib R4 + Ia (R3 + R4)))/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  V2 -&gt; (((Ia + Ib) R1 R2 + Ib (R1 + R2) R3) R4)/(
   R2 (R3 + R4) + R1 (R2 + R3 + R4)), 
  V3 -&gt; Ia R4 + Ib R4 - ((R1 + R2) R4 (Ib R4 + Ia (R3 + R4)))/(
    R2 (R3 + R4) + R1 (R2 + R3 + R4)) + Ib R5}}

In order to find the Thevenin equivalence, we need to take a loot at the open-circuit voltage \$\text{V}_\text{th}=\text{V}_1-\text{V}_2\$ (when \$\text{R}_3\to\infty\$) and at the short circuit current \$\text{I}_3\$ (when \$\text{R}_3\to0\$):
In[2]:=FullSimplify[
 Solve[{Ia == I1 + I5, I5 == I2 + I3, I4 == Ib + I3, I1 == Ia + I6, 
   0 == I2 + I6 + I7, I4 == Ib + I7, I1 == V1/R1, I2 == V1/R2, 
   I3 == (V1 - V2)/Infinity, I4 == V2/R4, Ib == (V3 - V2)/R5}, {I1, 
   I2, I3, I4, I5, I6, I7, V1, V2, V3}]]

Out[2]={{I1 -&gt; (Ia R2)/(R1 + R2), I2 -&gt; (Ia R1)/(R1 + R2), I3 -&gt; 0, I4 -&gt; Ib,
   I5 -&gt; (Ia R1)/(R1 + R2), I6 -&gt; -((Ia R1)/(R1 + R2)), I7 -&gt; 0, 
  V1 -&gt; (Ia R1 R2)/(R1 + R2), V2 -&gt; Ib R4, V3 -&gt; Ib (R4 + R5)}}

In[3]:=FullSimplify[(Ia R1 R2)/(R1 + R2) - Ib R4]

Out[3]=(Ia R1 R2)/(R1 + R2) - Ib R4

In[4]:=Limit[(Ia R1 R2 - Ib (R1 + R2) R4)/(R2 (R3 + R4) + R1 (R2 + R3 + R4)),
  R3 -&gt; 0]

Out[4]=(Ia R1 R2 - Ib (R1 + R2) R4)/(R2 R4 + R1 (R2 + R4))

So, using your values we get:

$$\text{V}_\text{th}=-\frac{11}{3}\approx-3.66667\space\text{V}\tag3$$
$$\text{I}_\text{th}=-\frac{11}{5}=-2.2\space\text{A}\tag4$$
$$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=\frac{5}{3}\approx1.66667\space\Omega\tag5$$