# 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?

• Notice that $|x-1||y+3|$ is an integer between $2$ and $5$, so there are only few possibilities to check.
– Sil
Mar 22, 2021 at 11:26
• Since $x$ is a negative integer, $|x - 1| \geq 2.$ This means that $|y+3|$ must be less than $3$. Mar 22, 2021 at 11:27
• thanks for such quick responses...but can you please direct me in the way of how I was trying to solve this rather than taking a whole new solution approach. If you have new approach...please explain down as answers...thanks! Mar 22, 2021 at 11:33
• In your second case we have $|y+3|=(-y-3)$ and also $1-x>1$ (in fact $1-x\geq 2$),and so by $(1-x)(-y-3) \le 5$ we must have $-y-3 \leq \frac{5}{2}$. This gives you a lower bound for $y$ similarly as in first case.
– Sil
Mar 22, 2021 at 11:51
• @Sil : can you please elaborate your last comment? I understood the part that $1-x \ge 2$ but i did not understood after that. If you kindly elaborate the later part of your comment. Mar 22, 2021 at 12:09

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.

• Can you please elaborate "We can make $|y+3|$ equal $1$ or $2$ twice though, with $y=-2,-4$ and $y=-1,-5$, so each one is doubled up to give $10$ solutions." ? Thanks ! Mar 22, 2021 at 11:51
• @Ganit; is that better?
– JMP
Mar 22, 2021 at 13:05

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.

$$10.$$

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. $$(x,y) = \{$$ $$(-1, -2),(-1, -4), (-2,-2),(-2,-4), (-1,-1),(-1,-5), (-3,-2)(-3,-4), (-4,-2),(-4,-4)$$ $$\}$$. • Shouldn't there be$2$in "So just rewrite them as$m,n$and we have$1 \le mn \le 5$."? Mar 22, 2021 at 15:28 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 $$10$$. • Can you please explain "We have since$x \le -1$,$|x-1|\ge 2,$we must have $$\frac22 \le |y+3|\le \frac52$$" as I am not able to understand this step? Are you dividing the in equality by$|x-1|$? Mar 23, 2021 at 3:00 • I made a msitake earlier, we can conclude that$|y+3| \le \frac{5}{|x+1|} \le \frac52\$. Mar 23, 2021 at 3:43