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.

y must be of the form $y=-(2k+1)5\geq-20$ which gives:
$(2k+1)5= (-5, -15) \geq -20$ $\rightarrow$ $2k+1=1, 3$
So for any $k\geq 0$ there will be an integer solution for x. in fact number of solutions is 2 when y is negative. Examples:
$2k+1=1$ $\rightarrow$ $y=-5>-20$, gives : $2x=25+99$ $\rightarrow$ $x=62$
$2k+1=3$ $\rightarrow$ $y=-15>-20$ ,gives: $2x=99+75$ $\rightarrow$ $x=87$
Now suppose x is negative then we must have:
$2x=-4, -14, -24, -34, -44, -54, -64, -74, -84, -94$
And corresponding y is:
$y=19, 17, 15, 13, 11, 9, 7, 5, 3, 1$
But this is not possible because dute to statement $x>y\geq-20$.
Now suppose x and y are both positive we have:
$x=47, 42, 37, 32, 27, 22,17, 12, 7, 2$
$y=1, 3, 5, 7, 9, 11, 13, 15, 17, 19 $
In this case only following solutions safices the condition:
$(x, y)=(47,1),(42, 3), (37, 5), (32, 7), (27, 9), (22, 11), (17, 13)$
Hence number of solutions is 9.
