Can you please help me solve this circuit?
Well, let's solve this mathematically. We have 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:
$$
\begin{cases}
0=\text{I}_1+\text{I}_2+\text{k}\cdot\text{I}_2=\text{I}_1+\text{I}_2\left(1+\text{k}\right)\\
\\
\text{I}_2=\text{I}_\text{x}+\text{I}_3\\
\\
0=\text{I}_\text{i}+\text{I}_3+\text{k}\cdot\text{I}_2\\
\\
\text{I}_\text{i}=\text{I}_\text{x}+\text{I}_1
\end{cases}\tag1
$$
When we use and apply Ohm's law, we can write the following set of equations:
$$
\begin{cases}
\text{I}_1=\frac{\text{V}_1}{\text{R}_1}\\
\\
\text{I}_2=\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_2-\text{V}_\text{i}}{\text{R}_3}
\end{cases}\tag2
$$
Substitute \$(2)\$ into \$(1)\$, in order to get:
$$
\begin{cases}
0=\frac{\text{V}_1}{\text{R}_1}+\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\cdot\left(1+\text{k}\right)\\
\\
\frac{\text{V}_1-\text{V}_2}{\text{R}_2}=\text{I}_\text{x}+\frac{\text{V}_2-\text{V}_\text{i}}{\text{R}_3}\\
\\
0=\text{I}_\text{i}+\frac{\text{V}_2-\text{V}_\text{i}}{\text{R}_3}+\text{k}\cdot\frac{\text{V}_1-\text{V}_2}{\text{R}_2}\\
\\
\text{I}_\text{i}=\text{I}_\text{x}+\frac{\text{V}_1}{\text{R}_1}
\end{cases}\tag3
$$
Now, we can solve for \$\text{V}_1\$:
$$\text{V}_1=\frac{\text{R}_1\left(1+\text{k}\right)\left(\text{V}_\text{i}-\text{I}_\text{x}\text{R}_3\right)}{\text{R}_1\left(1+\text{k}\right)+\text{R}_2+\text{R}_3}\tag4$$

Using your values, I got:
$$\text{V}_1=-\frac{35}{2}=-17.5\space\text{V}\tag5$$