Applying superposition: not enough equations when analyzing circuit with current sources

I am studying the superposition principle using Engineering Circuit Analysis by Hayt, Kemmerly, and Durbin. The book gives the following exercise:

PRACTICE 5.2: For the circuit of Fig. 5.7, use superposition to obtain the voltage across each current source.

My Attempt:

Using the superposition, I first deactivate the \$2\ \text{A}\$ current source.

Next, I apply KCL (Kirchhoff’s Current Law) at the red highlighted node:

$$\frac{v_1`-v_2`}{15\ \Omega}-i`+4\cdot i` =0\ \text{A} \label{1} \tag{I}$$

and I also use

$$i`=\frac{v_2`}{5\ \Omega} \label{2} \tag{II}$$

Substituting \eqref{1} into \eqref{2} I obtain

$$\frac{1}{15\ \Omega} \cdot v_1` + \frac{8}{15\ \Omega} \cdot v_2` = 0\ \text{A}$$

Where I’m stuck:

At this point, I end up with one equation involving two unknowns \$v_1`\$ and \$v_2`\$. I’m not sure how to obtain a second independent equation. A hint in the right direction would be greatly appreciated.

Thanks to the answers:

so first only open \$2\ \text{A}\$ current source. apply KCL on \$v_1`\$ and KCL on \$v_2`\$:

$$\begin{align} -\frac{22}{105\ \Omega}\cdot v_1` \ + \ \frac{1}{15\ \Omega}\cdot v_2` \ &= - \frac{3}{7}\ \text{A} \\ \frac{1}{15\ \Omega}\cdot v_1` \ + \ \frac{8}{15\ \Omega}\cdot v_2` \ &= 0\ \text{A} \\ \end{align}$$

Solution: \$v_1` \ =\ \frac{120}{61}\ \text{V} \$, \$v_2` \ =\ -\frac{15}{61}\ \text{V} \$

then only short \$3\ \text{V}\$ voltage source. apply KCL on \$v_1``\$ and KCL on \$v_2``\$:

$$\begin{align} -\frac{22}{105\ \Omega}\cdot v_1`` \ + \ \frac{1}{15\ \Omega}\cdot v_2`` \ &= - 2\ \text{A} \\ \frac{1}{15\ \Omega}\cdot v_1`` \ + \ \frac{8}{15\ \Omega}\cdot v_2`` \ &= 0\ \text{A} \\ \end{align}$$

Solution: \$v_1`` \ =\ \frac{560}{61}\ \text{V} \$, \$v_2`` \ =\ -\frac{70}{61}\ \text{V} \$