<s>[INST] 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 !
 [/INST] Both $|x-1|$ and $|y+3|$ are non-negative integers. As the product must lie between $2$ and $5$, the only candidates are:
$$(1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (3,1), (4,1), (5,1)$$
However, we can't force $|x-1|=1$ with a negative $x$, so this leaves just five possible solutions:
$$(2,1), (2,2), (3,1), (4,1), (5,1)$$
We can make $|y+3|$ equal $1$ or $2$ twice though, with $|-2
+3|=1,|-4+3|=1$ and similarly with $y=-1,-5$, so each pair can be made two times, once with each $y$, and this gives $10$ solutions.
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