I looked to see where the first equation would have (x,y) as integers:
it's true for (6,2).
Plugging those values into the second equation shows that it is a solution.
Lucky guess.
You can factor the second equation as:
√((x2)^2 + (y+1)^2) + √((x10)^2 + (y5)^2)
Plugging in (6,2):
√(16 + 9) + √(16 + 9) = 5+5 = 10
So, (6,2) is a solution.
Now, with a little (or a lot of) help from Wolfram Alpha:
Putting the second equation (without =10) into Wolfram Alpha shows that the z value is always >= 10.
Setting it equal to 10 shows a real solution to be:
2 <= x <= 10 and y = (1/4)(3x  10) or 3x4y = 10
3x + 4y = 26
3x  4y = 10
intersects at (6,2) so (I think) this is the only solution.

Posted by Larry
on 20230715 14:04:48 