How to find the value of alpha in this curcuit?
You have a \$9\:\text{A}\$ current source that leaves from the top end by only two paths and arrives to the bottom end via only two paths. You have two unknowns, \$\alpha\$ and \$I_x\$. So this feels convenient to me. Let's walk through the two paths using KVL to produce two equations. The first equation walks around the left side loop's KVL and the second equation walks around the right side loop's KVL.

$$\begin{align*}
\left(9\:\text{A}-I_x\right)\cdot 2\:\Omega + \left(9\:\text{A}-I_x +\alpha\:I_x\right)\cdot 2\:\Omega&amp;=V_{9\:\text{A}}\\\\
I_x\cdot 1\:\Omega+\left(I_x-\alpha\: I_x\right)\cdot 2\:\Omega&amp;=V_{9\:\text{A}}
\end{align*}$$

These are easily reduced to:

$$\begin{align*}
\left(2 -\alpha\right)\cdot I_x&amp;=18\:\text{A}-\frac{V_{9\:\text{A}}}{2\:\Omega}\\\\
\left(3-2\:\alpha\right)\cdot I_x&amp;=\frac{V_{9\:\text{A}}}{1\:\Omega}
\end{align*}$$

Which is easily solved simultaneously for \$\alpha\$ and for \$I_x\$.

Note that G36, in a comment I upped above, points out that you know something about the voltage difference across the \$9\:\text{A}\$ current source. Use that information to provide the value for \$V_{9\:\text{A}}\$.



Just look at the \$9\:\text{A}\$ source for a moment. It splits into two parts.


\$9\:\text{A}-I_x\$ goes to the left, through a \$2\:\Omega\$ resistor at the top left, where it then combines with an arriving current, \$\alpha\: I_x\$, that must be added to it before going through the remaining \$2\:\Omega\$ resistor at the bottom left. Clearly, the voltage drop across the first \$2\:\Omega\$ resistor is \$\left(9\:\text{A}-I_x\right)\cdot 2\:\Omega\$ and the voltage drop across the second \$2\:\Omega\$ resistor is \$\left(9\:\text{A}-I_x+\alpha\: I_x\right)\cdot 2\:\Omega\$.
\$I_x\$ goes to the right, through a \$1\:\Omega\$ resistor at the top right, splitting into two parts: \$\alpha\: I_x\$ and also therefore also \$I_x-\alpha\: I_x\$ that goes through a \$2\:\Omega\$ resistor at the bottom right. Clearly, the voltage drop across the \$1\:\Omega\$ resistor is \$I_x\cdot 1\:\Omega\$ and the voltage drop across the \$2\:\Omega\$ resistor is \$\left(I_x-\alpha\: I_x\right)\cdot 2\:\Omega\$.


You should be able to see why the current split occurs at the top end of the \$9\:\text{A}\$ current source such that \$I_x\$ goes to the right and \$9\:\text{A}-I_x\$ goes to the left. You should also clearly understand that the sum of the two currents arriving at the bottom end of the \$9\:\text{A}\$ current source must combine to make up that same \$9\:\text{A}\$ current source magnitude. There is no option, otherwise. The other current source, \$\alpha\: I_x\$, subtracts current from node and injects it into another node.

You merely need to remain consistent about the indicated directions in the schematic. If the current arrow says a current is leaving a node, then you subtract it. If the current arrow says a current is entering a node, then you add it. Do not try and impose some mental "sense" about what might "really" be happening. Your intuition is irrelevant, here. Just follow the directions indicated and assume they are correct. If they are NOT correct, then the sign will be negative and that will tell you that the current is actually going opposite to what you assumed from the indicated arrows. But that is fine. The point is to form up the equations in a consistent manner.