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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 !

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  • $\begingroup$ Notice that $|x-1||y+3|$ is an integer between $2$ and $5$, so there are only few possibilities to check. $\endgroup$
    – Sil
    Mar 22 at 11:26
  • $\begingroup$ Since $x$ is a negative integer, $|x - 1| \geq 2.$ This means that $|y+3|$ must be less than $3$. $\endgroup$ Mar 22 at 11:27
  • $\begingroup$ 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! $\endgroup$
    – Ganit
    Mar 22 at 11:33
  • $\begingroup$ 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. $\endgroup$
    – Sil
    Mar 22 at 11:51
  • $\begingroup$ @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. $\endgroup$
    – Ganit
    Mar 22 at 12:09
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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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  • $\begingroup$ 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 ! $\endgroup$
    – Ganit
    Mar 22 at 11:51
  • $\begingroup$ @Ganit; is that better? $\endgroup$
    – JMP
    Mar 22 at 13:05
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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.

$$10.$$

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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.

$(x,y) = \{$

$(-1, -2),(-1, -4), (-2,-2),(-2,-4), (-1,-1),(-1,-5), (-3,-2)(-3,-4), (-4,-2),(-4,-4)$

$\}$.

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  • $\begingroup$ Shouldn't there be $2$ in "So just rewrite them as $m,n$ and we have $1 \le mn \le 5$."? $\endgroup$
    – Ganit
    Mar 22 at 15:28
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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 $10$.

enter image description here

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  • $\begingroup$ 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 $|x-1|$? $\endgroup$
    – Ganit
    Mar 23 at 3:00
  • $\begingroup$ I made a msitake earlier, we can conclude that $|y+3| \le \frac{5}{|x+1|} \le \frac52$. $\endgroup$ Mar 23 at 3:43

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