That's correct.Now that I know that Vx=Vi*α the equation for Ix would then be:

Ix= (Vi-(Vi*α ))/R1=(Vi(1-α ))/R1

Using tex for readibility.

\( \small I_X = \frac{V_i-\alpha V_i}{R_1} \)

Well, yes. But Vo is an unknown.Vo should be Vx minus the voltage across R2 from left to right, shouldn't it?

Check your math again.That voltage (across R2) would be equal to Ix*R2 where R2=R1 so then Vi(1-α)

Therefore Vo=(Vi*α)-(Vi(1-α)=Vi*α-Vi-Vi*α=-Vi

That doesn't seem right?