First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).
Well, we are trying to analyze the following circuit:

simulate this circuit – Schematic created using CircuitLab
When we use and apply KCL, we can write the following set of equations:
$$
\begin{cases}
\text{I}_\text{a}=\text{I}_1+\text{I}_2+\text{I}_5\\
\\
\text{I}_3=\text{I}_2+\text{I}_6\\
\\
\text{I}_4=\text{I}_3+\text{I}_5\\
\\
\text{I}_6=\text{I}_4+\text{I}_7\\
\\
\text{I}_1=\text{I}_\text{a}+\text{I}_7
\end{cases}\tag1
$$
When we use and apply Ohm's law, we can write the following set of equations:
$$
\begin{cases}
\text{I}_1=\frac{\text{V}_1}{\text{R}_1}\\
\\
\text{I}_2=\frac{\text{V}_1-\text{V}_\text{i}}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_\text{i}-\text{V}_2}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_2}{\text{R}_4}\\
\\
\text{I}_5=\frac{\text{V}_1-\text{V}_2}{\text{R}_5}
\end{cases}\tag2
$$
We can subsitute \$(2)\$ into \$(1)\$, to get:
$$
\begin{cases}
\text{I}_\text{a}=\frac{\text{V}_1}{\text{R}_1}+\frac{\text{V}_1-\text{V}_\text{i}}{\text{R}_2}+\frac{\text{V}_1-\text{V}_2}{\text{R}_5}\\
\\
\frac{\text{V}_\text{i}-\text{V}_2}{\text{R}_3}=\frac{\text{V}_1-\text{V}_\text{i}}{\text{R}_2}+\text{I}_6\\
\\
\frac{\text{V}_2}{\text{R}_4}=\frac{\text{V}_\text{i}-\text{V}_2}{\text{R}_3}+\frac{\text{V}_1-\text{V}_2}{\text{R}_5}\\
\\
\text{I}_6=\frac{\text{V}_2}{\text{R}_4}+\text{I}_7\\
\\
\frac{\text{V}_1}{\text{R}_1}=\text{I}_\text{a}+\text{I}_7
\end{cases}\tag3
$$
Now, we can set up a Mathematica-code to solve for all the voltages and currents:
In[1]:=Clear["Global`*"];
FullSimplify[
Solve[{Ia == I1 + I2 + I5, I3 == I2 + I6, I4 == I3 + I5,
I6 == I4 + I7, I1 == Ia + I7, I1 == V1/R1, I2 == (V1 - Vi)/R2,
I3 == (Vi - V2)/R3, I4 == V2/R4, I5 == (V1 - V2)/R5}, {I1, I2, I3,
I4, I5, I6, I7, V1, V2}]]
Out[1]={{I1 -> (Ia R2 R4 R5 +
Ia R2 R3 (R4 + R5) + (R2 + R3) R4 Vi + (R3 + R4) R5 Vi)/(
R1 R2 R3 + R1 R2 R4 + R1 R3 R4 +
R2 R3 R4 + (R1 + R2) (R3 + R4) R5),
I2 -> (Ia R1 (R4 R5 +
R3 (R4 + R5)) - (R3 (R1 + R4) + (R3 + R4) R5) Vi)/(
R1 R2 R3 + R1 R2 R4 + R1 R3 R4 +
R2 R3 R4 + (R1 + R2) (R3 + R4) R5),
I3 -> (-Ia R1 R2 R4 + R1 (R2 + R5) Vi + R2 (R4 + R5) Vi)/(
R1 R2 R3 + R1 R2 R4 + R1 R3 R4 +
R2 R3 R4 + (R1 + R2) (R3 + R4) R5),
I4 -> (Ia R1 R2 R3 + R2 R5 Vi + R1 (R2 + R3 + R5) Vi)/(
R1 R2 R3 + R1 R2 R4 + R1 R3 R4 +
R2 R3 R4 + (R1 + R2) (R3 + R4) R5),
I5 -> (Ia R1 R2 (R3 + R4) + (R1 R3 - R2 R4) Vi)/(
R1 R2 R3 + R1 R2 R4 + R1 R3 R4 +
R2 R3 R4 + (R1 + R2) (R3 + R4) R5),
I6 -> (-Ia R1 ((R2 + R3) R4 + (R3 + R4) R5) + ((R2 + R3) (R1 +
R4) + (R1 + R2 + R3 + R4) R5) Vi)/(
R1 R2 R3 + R1 R2 R4 + R1 R3 R4 +
R2 R3 R4 + (R1 + R2) (R3 + R4) R5),
I7 -> (-Ia R1 (R3 R4 +
R2 (R3 + R4) + (R3 + R4) R5) + ((R2 + R3) R4 + (R3 +
R4) R5) Vi)/(
R1 R2 R3 + R1 R2 R4 + R1 R3 R4 +
R2 R3 R4 + (R1 + R2) (R3 + R4) R5),
V1 -> (Ia R1 R2 (R4 R5 + R3 (R4 + R5)) +
R1 ((R2 + R3) R4 + (R3 + R4) R5) Vi)/(
R1 R2 R3 + R1 R2 R4 + R1 R3 R4 +
R2 R3 R4 + (R1 + R2) (R3 + R4) R5),
V2 -> (R4 (Ia R1 R2 R3 + R2 R5 Vi + R1 (R2 + R3 + R5) Vi))/(
R1 R2 R3 + R1 R2 R4 + R1 R3 R4 +
R2 R3 R4 + (R1 + R2) (R3 + R4) R5)}}
Now, we can find:
- \$\text{V}_\text{th}\$ we get by finding \$\text{V}_1-\text{V}_2\$ and letting \$\text{R}_5\to\infty\$:
$$\text{V}_\text{th}=\frac{\text{I}_\text{a}\text{R}_1\text{R}_2\left(\text{R}_3+\text{R}_4\right)+\left(\text{R}_1\text{R}_3-\text{R}_2\text{R}_4\right)\text{V}_\text{i}}{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}\tag4$$
- \$\text{I}_\text{th}\$ we get by finding \$\text{I}_5\$ and letting \$\text{R}_5\to0\$:
$$\text{I}_\text{th}=\frac{\text{I}_\text{a}\text{R}_1\text{R}_2\left(\text{R}_3+\text{R}_4\right)+\left(\text{R}_1\text{R}_3-\text{R}_2\text{R}_4\right)\text{V}_\text{i}}{\text{R}_2\text{R}_3\left(\text{R}_1+\text{R}_4\right)+\text{R}_1\text{R}_4\left(\text{R}_2+\text{R}_3\right)}\tag5$$
- \$\text{R}_\text{th}\$ we get by finding:
$$\text{R}_\text{th}=\frac{\text{V}_\text{th}}{\text{I}_\text{th}}=\frac{\text{R}_2\text{R}_3\left(\text{R}_1+\text{R}_4\right)+\text{R}_1\text{R}_4\left(\text{R}_2+\text{R}_3\right)}{\left(\text{R}_1+\text{R}_2\right)\left(\text{R}_3+\text{R}_4\right)}\tag6$$
Where I used the following Mathematica-codes:
In[2]:=FullSimplify[
Limit[(((Ia R1 R2 (R4 R5 + R3 (R4 + R5)) +
R1 ((R2 + R3) R4 + (R3 + R4) R5) Vi)/(
R1 R2 R3 + R1 R2 R4 + R1 R3 R4 +
R2 R3 R4 + (R1 + R2) (R3 + R4) R5)) - ((
R4 (Ia R1 R2 R3 + R2 R5 Vi + R1 (R2 + R3 + R5) Vi))/(
R1 R2 R3 + R1 R2 R4 + R1 R3 R4 +
R2 R3 R4 + (R1 + R2) (R3 + R4) R5))), R5 -> Infinity]]
Out[2]=(Ia R1 R2 (R3 + R4) + (R1 R3 - R2 R4) Vi)/((R1 + R2) (R3 + R4))
In[3]:=FullSimplify[
Limit[(Ia R1 R2 (R3 + R4) + (R1 R3 - R2 R4) Vi)/(
R1 R2 R3 + R1 R2 R4 + R1 R3 R4 + R2 R3 R4 + (R1 + R2) (R3 + R4) R5),
R5 -> 0]]
Out[3]=(Ia R1 R2 (R3 + R4) + (R1 R3 - R2 R4) Vi)/(
R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4)
In[4]:=FullSimplify[%2/%3]
Out[4]=(R1 R2 R3 + R2 R3 R4 + R1 (R2 + R3) R4)/((R1 + R2) (R3 + R4))
So, using your values we get:
- $$\text{V}_\text{th}=\frac{29028}{4175}\approx6.95281\space\text{V}\tag7$$
- $$\text{I}_\text{th}=\frac{16933}{13736000}\approx0.00123275\space\text{A}\tag8$$
- $$\text{R}_\text{th}=\frac{6593280}{1169}\approx5640.1\space\Omega\tag9$$