This was already mentioned by Russel, but I hope to present it in a different way.
The main problem here, it seems to me, is that your book (or whatever source of information you're using) missed one important point: The voltage between inverting and non-inverting inputs of an ideal operational amplifier should always be zero with this and similar setups. If we include that assumption and take a look at the circuit, we can get a logical answer.
The output of an op-amp is modelled as an ideal controlled voltage source. The input impedance is infinite and no current flows into the op-amp. So far so good. Next, we know that the voltage between the inputs is zero, so we know that the voltage with respect to ground and the inverting input is same as the one on the non-inverting input. That voltage comes from the ideal controlled voltage source at the output. Next, let's take a look at the current issue. Since we have infinite input impedance, no current flows into the operational amplifier, so from where does the output current come? Well from the ideal controlled voltage source at the output.
As I said, the voltage source is ideal, so it can source infinite current, it's controlled so you have your gain, the current is set by resistor and there's no contradiction there at all. In reality, the current will come from the power supply pins and be limited by construction of the operational amplifier, but this is a mathematical model. So let's take a look at a pretty pictures now:
The first image may seem a bit drastic:
I've crossed out the op-amp on purpose here. It seems to me that trees are obstructing your view of the forest here. If we remove the op-amp symbol and take a look at how we're supposed to model it instead (note the \$ 100 \mbox{ }G \Omega \$ resistor):
We can clearly see that the current is coming from the one terminal voltage source which is the output of the op-amp.
Next, I'll show a bit more complex version of the same circuit and explain how it degenerates into what you've shown:
Let's see what we can see here:
We've got the input voltage \$U_i\$, the output voltage \$U_o\$ and the resistors \$R_1\$ and \$R_2\$.
Now we know from our model that the voltage between the inputs is zero, so we can write following safely: \$U_i-R_1I=0\$, since the resistor \$R_1\$ has a short circuit to inverting input. From that we get the current: \$I=\frac{U_i}{R_1}\$. The current can only come from the op-amp output in this case, so we know that it is the current going through the resistor \$R_2\$ too. From that we get the equation for the output voltage of the op-amp: \$ U_o-R_2I-U_i=0\$ and after that: \$U_o=R_2 I + U_i= R_2 \frac{U_i}{R_1} + U_i=U_i(\frac{R_2}{R_1}+1)\$. From this, we have \$ \frac{U_o}{U_i}=1+ \frac{R_2}{R_1}\$. In the circuit you showed, equivalent elements would be \$R_2=0\$ and \$R_1=\infty\$. As you can see, the output current isn't a problem with this setup and again, there's no contradiction here.
With the few assumptions I've shown and few equations, you can do basic op-amp circuits without any problems. I recommend that you read from freely downloadable books Amplifiers and Bits: An Introduction to Selecting Amplifiers for Data Converters pages 6 and 7 and from Op Amps for Everyone Design Guide chapter 3 (or at least take a good look at the pictures there). Both books (well, a book and an application report) are by Texas Instruments (a major op-amp manufacturer) and should come up on most popular search engines as the first response.
KCL at the center node gives:
$$ \frac{-V_x-50}{10} + 5 + 0.1V_x + \frac{-V_x}{20} = 0 $$
Which is only an equation of \$V_x\$ and can be immediately solved to get \$V_x = 0\$.
KCL at the top node (with voltage \$V_T\$):
$$ \frac{V_T - 50}{20} - 5 + I_Y = 0 $$
Ohm's law on the 25 ohm resistor gives \$ I_Y = \frac{V_T - 5I_Y}{25} \implies V_T = 30I_Y\$. Using this in the preceeding equation gives:
$$ \frac{30I_Y - 50}{20} - 5 + I_Y = 0 \implies I_Y = 3 \text{A}$$
So, along with your own answer and Alfred Centauri's answer, we can conclude that your book's answer of \$V_X = 70 V\$ is incorrect.
Best Answer
There are hidden power supplies connecting from the ground node to the positive and negative supply pins of the op-amp. Current to the output pin of the op-amp flows from these supplies.
I'll redraw the circuit including these supplies:
simulate this circuit – Schematic created using CircuitLab
Now can you see how current from the op-amp output can flow back to the op-amp's positive supply pin?