Open circuit voltage understanding
Well, we are 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:
$$\text{I}_2=\text{I}_1+\text{I}_3\tag1$$
When we use and apply Ohm's law, we can write the following set of equations:
$$
\begin{cases}
\text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\
\\
\text{I}_2=\frac{\text{V}_1}{\text{R}_2}\\
\\
\text{I}_3=\frac{0-\text{V}_2}{\text{R}_3}
\end{cases}\tag2
$$
We also know that \$\text{V}_1-\text{V}_2=\text{V}_\text{n}\$.
Now, it is not hard to solve for \$\text{V}_2\$ and \$\text{I}_1\$:

$$\text{V}_2=\frac{\text{R}_2\text{R}_3\text{V}_\text{i}-\text{R}_3\left(\text{R}_1+\text{R}_2\right)\text{V}_\text{n}}{\text{R}_2\text{R}_3+\text{R}_1\left(\text{R}_2+\text{R}_3\right)}\tag4$$
$$\text{I}_1=\frac{\left(\text{R}_2+\text{R}_3\right)\text{V}_\text{i}-\text{R}_2\text{V}_\text{n}}{\text{R}_2\text{R}_3+\text{R}_1\left(\text{R}_2+\text{R}_3\right)}\tag5$$

Where I used Mathematica-code to solve for that:
In[1]:=FullSimplify[
 Solve[{I2 == I1 + I3, I1 == (Vi - V1)/R1, I2 == V1/R2, 
   I3 == (0 - V2)/R3, V1 - V2 == Vn}, {I1, I2, I3, V1, V2}]]

Out[1]={{I1 -&gt; ((R2 + R3) Vi - R2 Vn)/(R2 R3 + R1 (R2 + R3)), 
  I2 -&gt; (R3 Vi + R1 Vn)/(R2 R3 + R1 (R2 + R3)), 
  I3 -&gt; (-R2 Vi + (R1 + R2) Vn)/(R2 R3 + R1 (R2 + R3)), 
  V1 -&gt; (R2 (R3 Vi + R1 Vn))/(R2 R3 + R1 (R2 + R3)), 
  V2 -&gt; (R2 R3 Vi - (R1 + R2) R3 Vn)/(R2 R3 + R1 (R2 + R3))}}


Now, we can apply the theory from above on your circuit. We know that \$\text{V}_\text{n}:=\text{k}\cdot\left(-\text{I}_1\right)\$, so we can rewrite equation \$(4)\$ as follows:
$$\text{V}_2=\frac{\text{R}_3\left(\text{k}+\text{R}_2\right)\text{V}_\text{i}}{\text{R}_2\text{R}_3+\text{R}_1\left(\text{R}_2+\text{R}_3\right)-\text{k}\text{R}_2}\tag6$$
Where I used the following code:
In[2]:=FullSimplify[
 Solve[{I2 == I1 + I3, I1 == (Vi - V1)/R1, I2 == V1/R2, 
   I3 == (0 - V2)/R3, V1 - V2 == -k*I1}, {I1, I2, I3, V1, V2}]]

Out[2]={{I1 -&gt; ((R2 + R3) Vi)/(-k R2 + R2 R3 + R1 (R2 + R3)), 
  I2 -&gt; ((-k + R3) Vi)/(-k R2 + R2 R3 + R1 (R2 + R3)), 
  I3 -&gt; ((k + R2) Vi)/(k R2 - R2 R3 - R1 (R2 + R3)), 
  V1 -&gt; (R2 (k - R3) Vi)/(k R2 - R2 R3 - R1 (R2 + R3)), 
  V2 -&gt; ((k + R2) R3 Vi)/(-k R2 + R2 R3 + R1 (R2 + R3))}}

For your circuit we must compute \$\text{V}_2\$ when \$\text{R}_3\to\infty\$, so we get:
$$\text{V}_\text{out}:=\lim_{\text{R}_3\to\infty}\text{V}_2=\frac{\left(\text{k}+\text{R}_2\right)\text{V}_\text{i}}{\text{R}_1+\text{R}_2}\tag7$$
Where I used the following code:
In[3]:=FullSimplify[
 Limit[((k + R2) R3 Vi)/(-k R2 + R2 R3 + R1 (R2 + R3)), 
  R3 -&gt; Infinity]]

Out[3]=((k + R2) Vi)/(R1 + R2)


So, in your circuit we get:
$$\text{V}_\text{out}=\frac{\left(20\cdot10^3+10\cdot1000\right)\cdot75}{40\cdot1000+10\cdot1000}=45\space\text{V}\tag8$$