Solve simultaneously x^x + y^x= 13 y^y - x^y = 19

2^2 + 3^2 = 13
3^3 - 2^3 = 19

x = 2, y =3
just trial and error :-)

by the looks of things,
x can't be equal to y
x and y must be > 1
===> if x< 1 then things don't work; similarly if y<1 things don't work
and x and y must be < 4
===> if x > 4 and y > 1, then equalities don't hold
===> if x > 1 and y > 4, then equalities don't hold
So, only possibilities are x is one of 2 or 3 and y is one of 2 or 3
And y > x
so, x = 2 and y = 3

Not particularly mathematical reasoning, but it works (almost a diophantine equation set)!

x^x + y^x= 13 => y = (13 - x^x)^(1/x)
y^y - x^y = 19
[(13 - x^x)^(1/x))^((13 - x^x)^(1/x)] - x^[(13-x^x)^(1/x)] = 19
x = 2
y = 3