Number of possible solutions to $2 \le |x-1||y+3| \le 5$, where $x$ and $y$ are negative integers 
If $$2 \le |x-1||y+3| \le 5$$ and both $x$ and $y$ are negative integers, find the number of possible combinations of $x$ and $y$ .

Below is my solution approach :-
As $x$ is a negative integer, hence $|x-1|$ in the $2 \le |x-1||y+3| \le 5$ will be come $-(x-1)$ or $(1-x)$ and the main equation will transform into $2 \le (1-x)|y+3| \le 5$.
$1st$ case when $y+3 \ge 0 \Rightarrow y \ge -3 \Rightarrow y \in \{{-3,-2,-1}\} $ as $y$ is a negative integer :
For $y=-3$ we can see that there is no valid solution for $x$ as $|y+3|$ part will become $0$, hence this case is invalid.
For $y=-2$ we get the solution set for $x$ to be $x \in \{{-4,-3,-2,-1}\} $  and total number of solutions possible in this case is $4$.
For $y=-1$ we get the solution set for $x$ to be $x \in \{{-1}\} $ and total number of solutions possible in this case is $1$.
So for this $1st$ case when $y+3 \ge 0$ we have in total 5 solutions.
$2nd$ case when $y+3 \lt 0 \Rightarrow y \lt -3$ and in this case $y$ will have infinite values and I am not able to proceed from here.

The answer for the total number of solutions provided is $10$ and you can see that I've been able to find out the $5$ solutions in my $1st$ case. Can someone please guide or help me about how to proceed in the 2nd case?

Thanks in advance !

It is not true that $y$ can take infinitely many numbers.
We have  since $x \le -1$, $|x-1|\ge 2,  $
$$\frac1{|x-1|}\le \frac12$$
we must have
$$|y+3| \le \frac{5}{|x+1|}\le \frac52$$
If $y< -3$, then we have $$-y-3 \le \frac52 $$
$$-\frac{11}2 \le y $$
Hence $-5 \le y < -3$, that is $-5 \le y \le -4$.
For $y=-4=-3-1$, by symmetry, the same $x$ for which $y=-3+1=-2$ satisfies the inequality would work,  there are $4$ of them.
For $y=-5=-3-2$, by symmetry, the same $x$ for which $y=-3+2=-1$ satisfies the inequlaity would work, there is $1$ of them.
Hence in total, there are $10$.

