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?
I'll show you an easy way using mesh analysis plus one node. First look at the node at the top of the right-hand side \$2\Omega\$ resistor:
$$i_1+2A=i_1+i_3$$
from which you get immediately \$i_3=2A\$. Then you need two more equations for \$i_1\$ and \$i_2\$. Use the outer mesh and note that the current through the left-hand \$2\Omega\$ resistor is \$i_1+i_2+2A\$:
$$(i_1+i_2+2A)\cdot 2\Omega + i_1\cdot 4\Omega+2A (=i_3)\cdot 2\Omega=10V\tag{1}$$
The second equation comes from the mesh in the lower left of the circuit:
$$(i_1+i_2+2A)\cdot 2\Omega+i_2\cdot 1\Omega=10V\tag{2}$$
Solving (1) and (2) for \$i_1\$ and \$i_2\$ gives the desired result.
Best Answer
I get the following four equations:
$$\begin{align*} 0\:\text{V}+24\:\text{V}-5\:\Omega\cdot\left(I_1-I_3\right)-V_{4\text{A}}-3\:\Omega\cdot 4\:\text{A}&=0\:\text{A} \\\\ 0\:\text{V}+3\:\Omega\cdot 4\:\text{A}+V_{4\text{A}}-10\:\Omega\cdot\left(I_2-I_3\right)-20\:\Omega\cdot I_2&=0\:\text{A} \\\\ 0\:\text{V}-5\:\Omega\cdot\left(I_3-I_1\right)-5\:\Omega\cdot I_3-10\:\Omega\cdot\left(I_3-I_2\right)&=0\:\text{A} \\\\ I_1-I_2&=4\:\text{A} \end{align*}$$
where \$V_{4\text{A}}\$ is the voltage across the current source.
I used the + sign on the top node and - on the bottom node of the current source. As shown below:
No need for any super- anything.
These solve out (using sympy):
Check your work against mine to see if you agree. Then check all this against the stated solution.
Toss the textbook into the trash can. And with it, the 8th edition of "Electronic Principles" by Malvino & Bates. And probably dozens of other terrible books on the topic of electronics. I've no idea how they get so bad, but they do. Even after 8 editions! The world is flush with garbage. I'm guessing it has to do, in part, with grad students. But I'm not sure.