Finding the solution set of the equation $\frac{2(2x+3)}{x+1}-\frac{2}{x+1}+3=0,x\in\mathbb R$

Question

Question

Which of the following is an empty set?

A) The set of prime numbers which are even.

B) The solution set of the equation $\frac{2(2x+3)}{x+1}-\frac{2}{x+1}+3=0,x\in\mathbb R$

C) $(A\times B)\cap(B\times A),$ where A and B are disjoint.

D) The set of reals which satisfy $x^2+ix+i-1=0$

(The answer given is C))

My Attempt:

I think the answer should be B), C), D).

Regarding option B)

$$\frac{4x+6-2}{x+1}+3=0\\4+3=0$$

Regarding D)

$x=\frac{-i\pm\sqrt{-1-4i+4}}2$

Please confirm.

Answer 1

A) is obviously nonempty since $2$ is in the set.

B) $x=-1$ is clearly not acceptable as a solution, since the equation is not defined at $x=-1$. Now assuming $x\neq-1$. Then we get $7x=-7$ which gives $x=-1$, contradiction. Therefore here we have an empty set.

C)If we assume that there is an $(a,b)$ in $(AXB)\bigcap(BXA)$ then $(a,b)$ is in $AXB$ and in $(BXA)$ i.e $a$ is in $A$ and $B$, contradiction. Here the set is empty

D)The set is nonempty since for $x=-1$ the equation is satisfied!

So I only disagree with the answer in (B)!!