Finding voltage in series-parallel

it appears to me that the current generated by 35 V and 2vx will collide each other

It may be that you are assuming that a voltage source, whether independent or controlled, must source current, i.e., supply power to the circuit.

But, at least in ideal circuit theory, there's nothing "wrong" with a voltage source sinking current, i.e., receiving power from the circuit.

For a real world example, consider that, when a battery is being charged, the current is in the opposite direction than when the battery is being discharged.

I would like to know how the current flows across 5 Ω resistor.

If you're planning to be an EE, don't write or say things like "current across"; current is through, voltage is across.

Now, this circuit is very easy to solve. There are two unknowns so you need two independent equations.

For the 1st, write a KVL equation clockwise 'round the loop:

$$35V = v_x + 2v_x - v_o \rightarrow 3v_x = 35V + v_o$$

Now, you need one more independent equation. Can you find one?

If you are connecting a resistor in series with 1 or more LEDs do this:

You will need to know the voltage drop beetwen LEDs (normally around 2V per LED) and the voltage input. The voltage that gonna drop beetwen your resistor is what didn't drop in the LEDs. So:

$$V_{in} = V_r + V_{led}$$

$$V_r = V_{in} - V_{led}$$

Then the current going thought the resistor gonna be the current you want to flow in the LEDs, most of the time I use 10mA, the max the LED can handle is normally around 20mA. That will determine how much bright your LED gonna be.

With your resistor voltage and current you can use ohms law to calculate its resistance.

$$V=R \cdot I$$

One thing to pay attention is the power dessipated in the resistor.

$$P=V_r\cdot I_r$$

If you have parallel LED association I would consider using this to calculate a resistor for each parallel. The problem with using only one resistor is that all the current from each LED will go though this resistor, so it may need to have a better power rate. Even if you get a resistor that has a good power rate you may consider calculating separated LED resistors because power rate won't prevent it from heating up.