You are right in that $V_x$ can't be 10 V. Fortunately, 10V for $V_x$ simplifies the circuit considerably, making it particularly easy to analyze.
Since $V_x$ is the same voltage as the supply on the left, both ends of the 5/11 Ω resistor are at the same voltage. That means the current thru it is zero. That means the current thru anything directly in series with is must also be zero. Therefore, the current thru the 5/11 Ω resistor, the 10 V power supply, and the 1 Ω resistor at right are all 0. The voltage across any resistor with 0 current thru it must also be 0. We therefore know that point A is at 0 V.
Taking the loop with the 2 V supply and the two 1 Ω resistors in isolation, we can easily see that 1 A must be circulating, as indicated by your curved arrow. 1 A thru a 1 Ω resistor causes a 1 V drop, so the right end of this loop must be 1 V higher than the left.
Putting this back into the rest of the circuit that is known to carry 0 current just doesn't work. According to previous logic, point A must be at 0 V, but adding the +2 V of the supply and -1 V of the resistor to the 10 V at $V_x$ yields 11 V at A, which is a contradiction.
Clearly something is wrong somewhere.