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 !
 A: 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.
A: The possible values of the product are $2,3,4,5$.
Hence the possible factorizations,
$$1\cdot2,2\cdot1,1\cdot3,3\cdot1,1\cdot4,2\cdot2,4\cdot1,1\cdot5,5\cdot1.$$
But $|x-1|\ge2$, which leaves
$$2\cdot1,3\cdot1,2\cdot2,4\cdot1,5\cdot1,$$
while $|y+3|$ can achieve $1$ and $2$ in two ways.

$$10.$$

A: If $x,y$ are integers then $|x-1|$ and $|y+3|$ are non negative integers.
So just rewrite them as $m,n$ and we have $1 \le mn \le 5$.
Those are small and simple numbers.
We can have $nm = 1,2,3,4,5$
But you said $x,y$ are both *negative integers.  So $x \le -1$ and so $x -1 \le -2$ so $|x- 1| \ge 2$.
so we can have:
$nm = 2$ and $|x-1| =2$ and $|y + 3| = 1$ so $x-1= -2$ and $x =-1$ while $|y+3| = 1$ so $y+3 = \pm 1$ so $y = 1-3=-2$ or $y = -1 -3 = -4$.
or
$nm =3$ and $|x-1| = 3$ and $|y+3| = 1$ so $x-1 =-3$ and $x = -2$ while, as above, y=-2, -4$.
or $nm = 4$ and either $|x-1|=2$ and $|y+3|=2$.  Or $|x-1| =4$ and $|y+3| =1$.
If $|x-1|=|y+3|= 2$ then $x = -1$ and $y+3 =\pm 2$ so $y = -1; -5$.
And if $|x-1|=4$ and $|y+3| =1$ then $x-1 = -4$ and $x =-3$ and $y= -2,4$ (as above).
And if $nm = 5$ then $|x-1| = 5$ so $x-1 = -5$ and $x =-4$ and $|y+3| = 1$ and $y=-2,-4$.
So all solutions are.
$(x,y) = \{$
$(-1, -2),(-1, -4), (-2,-2),(-2,-4), (-1,-1),(-1,-5), (-3,-2)(-3,-4), (-4,-2),(-4,-4)$
$\}$.
A: 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 $10$.

