<s>[INST] The number of pairs of integers $(x,y)$ satisfying $x ≥ y ≥ -20$ and $2x + 5y = 99$ is I tried to solve this question but I was unable to think how to get the number of integer pairs satisfying these conditions. Till now I have broken down this into :-
$2x+5y = 99$
==>$y=(99-2x)/5$
$y\geq-20$
==> $(99-2x)/5 \geq20$
==> $x\leq 99.5$
$x\geq y$
==> $x \geq (99-2x)/5$
==> $x\geq 99/7$
From here I am not able to process further to think of the way that will give me the solution.
 [/INST] $2x + 5y = 99$ is very rare and a strong restriction.
If $2a + 5b=99$ is a solution to find another solution $2(a\pm d) +5(b\mp e) =99=2a+5b$ we must have $2d - 5e = 0$ and that requires that $d$ be divisible by $5$ and $e$ be divisble by $2$.
so if $2a + 5b =99 $ is one answer than any $2(a + 5k) + 5(b-2k) =99$ is also a solution and these are the only solutions.
Now lets find our first solution.
$2x +5y = 99$ so $5y = 99-2x$ and we need the RHS to be divisible by $5$.  We can do that by letting $x = 2$ and $5y = 95$ so $y = 19$.
So if $2x + 5y = 99$ then $x = 2+5k$ and $y = 19 - 2k$ for some integer $k$ and for any integer $k$ $x = 2+5k$ and $y =19-2k$ will be a solution.
So we must have $2+5k \ge 19-2k\ge -20$.  How many integers key can that be true for?
We must have $2+5k \ge 19-2k$ so $7k \ge 17$ so $k \ge 3$.  And we have $19-2k \ge -20$ so $2k \le 39$ so $k \le 19$.  So that's it.
$3\le k \le 19$ and $x = 2+5k$ and $y = 19 - 2k$ are all the solution pairs.  There are $17$ of them:
$(x,y) = (17, 13),(22,11),(27,9).....,(97, -19)$.
All other solutions will either have $y > x$ (such as $(12,15)$) of have $y < -20$ (such as $(102, -21)$.
</s>